Question

# Write each function in standard form .

F(x)= -2(x-9)^{2} +15

Hint:

### The standard quadratic form is ax2+bx+c=y, the vertex form of a quadratic equation is y=a(x−h)2+k.

## The correct answer is: f(x) = -2x2 + 36x - 147.

### We have given the function in Vertex form , we have to convert it into standard form**.**

F(x)= -2(x-9)^{2} +15

On further solving this function we will get

F(x) = -2 ( x^{2} - 18x + 81 ) + 15

= -2x^{2} + 36x - 162 + 15

= -2x^{2} + 36x - 147

Therefore, the standard form of given function is f(x) = -2x^{2} + 36x - 147.

F(x)= -2(x-9)

^{2}+15

On further solving this function we will get

F(x) = -2 ( x

^{2}- 18x + 81 ) + 15

= -2x

^{2}+ 36x - 162 + 15

= -2x

^{2}+ 36x - 147

Therefore, the standard form of given function is f(x) = -2x

^{2}+ 36x - 147.

**Related Questions to study**

**Maths-**### Write each function in standard form . F(x)= 4(x+1)

Hint :- The standard quadratic form is ax

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 4(x+1)

On further solving this function we will get

F(x) = 4 ( x

= 4x

= 4x

Therefore, the standard form of given function is f(x) = 4x

Q39:

Complex Level : Hard

Blooms Level: Understanding

Write each function in standard form . F(x)= 0.1 (x-2)

Hint :- The standard quadratic form is ax

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 0.1(x-2)

On further solving this function we will get

F(x) = 0.1 ( x

= 0.1x

= 0.1x

Therefore, the standard form of given function is f(x) =0.1x### Write each function in standard form . F(x)= 4(x+1)

Maths-GeneralHint :- The standard quadratic form is ax

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 4(x+1)

On further solving this function we will get

F(x) = 4 ( x

= 4x

= 4x

Therefore, the standard form of given function is f(x) = 4x

Q39:

Complex Level : Hard

Blooms Level: Understanding

Write each function in standard form . F(x)= 0.1 (x-2)

Hint :- The standard quadratic form is ax

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 0.1(x-2)

On further solving this function we will get

F(x) = 0.1 ( x

= 0.1x

= 0.1x

Therefore, the standard form of given function is f(x) =0.1xMaths-### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

We have given two functions f(x) and h(x).

h(x) = x

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In h(x) = x

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

Maths-GeneralWe have given two functions f(x) and h(x).

h(x) = x

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In h(x) = x

Maths-### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

Solution:- We have given two functions f(x) and g(x).

g(x) = 2x

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In g(x)= 2x

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

Maths-GeneralSolution:- We have given two functions f(x) and g(x).

g(x) = 2x

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In g(x)= 2x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= = 4x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= = 4x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 5x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 5x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 0.4x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 0.4x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

Maths-### Find the axis of symmetry, vertex and y-intercept of the function

This quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 2x

### Find the axis of symmetry, vertex and y-intercept of the function

Maths-GeneralThis quadratic function is in standard form, f(x)=ax

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 2x

Maths-### Find the y-intercept of the following function

We have given a function

f(x) = -0.5x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = -0.5(0)

y = 2

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 2.### Find the y-intercept of the following function

Maths-GeneralWe have given a function

f(x) = -0.5x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = -0.5(0)

y = 2

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 2.Maths-### Find the y-intercept of the following function

We have given a function

f(x) = x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = (0)

y = 3

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 3.### Find the y-intercept of the following function

Maths-GeneralWe have given a function

f(x) = x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = (0)

y = 3

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 3.Maths-### Find the y-intercept of the following function

We have given a function

f(x) = 3x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 3(0)

y = 5

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 5.### Find the y-intercept of the following function

Maths-GeneralWe have given a function

f(x) = 3x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 3(0)

y = 5

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 5.Maths-### Find the y-intercept of the following function

We have given a function

f(x) = 2x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 2(0)

y = -7

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is -7.### Find the y-intercept of the following function

Maths-General We have given a function

f(x) = 2x

We will compare the given equation with the standard equation f(x)=ax

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 2(0)

y = -7

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is -7.

### Write each function in standard form . F(x)= 4(x+1)^{2} -3

^{2}+bx+c=y, the vertex form of a quadratic equation is y=a(x−h)

^{2}+k.

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 4(x+1)

^{2}-3

On further solving this function we will get

F(x) = 4 ( x

^{2}+ 2x + 1 ) – 3

= 4x

^{2}+ 8x + 4 – 3

= 4x

^{2}+ 8x + 1

Therefore, the standard form of given function is f(x) = 4x

^{2}+ 8x + 1.

Q39:

Complex Level : Hard

Blooms Level: Understanding

Write each function in standard form . F(x)= 0.1 (x-2)

^{2}– 0.1

Hint :- The standard quadratic form is ax

^{2}+bx+c=y, the vertex form of a quadratic equation is y=a(x−h)

^{2}+k.

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 0.1(x-2)

^{2}- 0.1

On further solving this function we will get

F(x) = 0.1 ( x

^{2}- 4x + 4 ) – 0.1

= 0.1x

^{2}– 0.4x + 0.4 – 0.1

= 0.1x

^{2}-0.4x + 0.3

Therefore, the standard form of given function is f(x) =0.1x

^{2}-0.4x + 0.3.

### Write each function in standard form . F(x)= 4(x+1)^{2} -3

^{2}+bx+c=y, the vertex form of a quadratic equation is y=a(x−h)

^{2}+k.

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 4(x+1)

^{2}-3

On further solving this function we will get

F(x) = 4 ( x

^{2}+ 2x + 1 ) – 3

= 4x

^{2}+ 8x + 4 – 3

= 4x

^{2}+ 8x + 1

Therefore, the standard form of given function is f(x) = 4x

^{2}+ 8x + 1.

Q39:

Complex Level : Hard

Blooms Level: Understanding

Write each function in standard form . F(x)= 0.1 (x-2)

^{2}– 0.1

Hint :- The standard quadratic form is ax

^{2}+bx+c=y, the vertex form of a quadratic equation is y=a(x−h)

^{2}+k.

Solution :- We have given the function in Vertex form , we have to convert it into standard form.

F(x)= 0.1(x-2)

^{2}- 0.1

On further solving this function we will get

F(x) = 0.1 ( x

^{2}- 4x + 4 ) – 0.1

= 0.1x

^{2}– 0.4x + 0.4 – 0.1

= 0.1x

^{2}-0.4x + 0.3

Therefore, the standard form of given function is f(x) =0.1x

^{2}-0.4x + 0.3.

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

h(x) = x^{2} + x – 3.5

h(x) = x

^{2}+ x – 3.5

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In h(x) = x

^{2}+ x – 3.5, a= 1, b= 1, and c = -3.5. So, the equation for the axis of symmetry is given by

x = −(1)/2(1)

x = -1/2

x = -0.5

The equation of the axis of symmetry for h(x) = x^{2} + x – 3.5 is x = -0.5.

The x coordinate of the vertex is the same:

h =-0.5

The y coordinate of the vertex is :

k = f(h)

k = h^{2} + h – 3.5

k = (-0.5)^{2} + (-0.5) – 3.5

k = 0.25 – 0.5 – 3.5

k = -3.75

Therefore, the vertex is (-0.5 , -3.75)

The minimum value of g(x) will be the y-coordinate of vertex = -3.75

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

h(x) = x^{2} + x – 3.5

h(x) = x

^{2}+ x – 3.5

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In h(x) = x

^{2}+ x – 3.5, a= 1, b= 1, and c = -3.5. So, the equation for the axis of symmetry is given by

x = −(1)/2(1)

x = -1/2

x = -0.5

The equation of the axis of symmetry for h(x) = x^{2} + x – 3.5 is x = -0.5.

The x coordinate of the vertex is the same:

h =-0.5

The y coordinate of the vertex is :

k = f(h)

k = h^{2} + h – 3.5

k = (-0.5)^{2} + (-0.5) – 3.5

k = 0.25 – 0.5 – 3.5

k = -3.75

Therefore, the vertex is (-0.5 , -3.75)

The minimum value of g(x) will be the y-coordinate of vertex = -3.75

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

g(x) = 2x^{2} + 8x + 3

g(x) = 2x

^{2}+ 8x + 3

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In g(x)= 2x

^{2}+ 8x + 3, a= 2, b= 8, and c=3. So, the equation for the axis of symmetry is given by

x = −(8)/2(2)

x = -8/4

x = -2

The equation of the axis of symmetry for g(x)= 2x^{2} + 8x + 3 is x = -2.

The x coordinate of the vertex is the same:

h =-2

The y coordinate of the vertex is :

k = f(h)

k = 2h^{2} + 8h + 3

k = 2(-2)^{2} + 8(-2) + 3

k = 8 - 16 + 3

k = -5

Therefore, the vertex is (-2 , -5)

The minimum value of g(x) will be the y-coordinate of vertex = -5

### Compare each function to f, shown in the table. Which function has lesser minimum value? Explain

g(x) = 2x^{2} + 8x + 3

g(x) = 2x

^{2}+ 8x + 3

For f(x) , minimum value of the function will be the y-coordinate of the given point which has minimum values

(1,0) , (2,-3) , (3,-4) , (4,-3) (5,0)

In the given points minimum value of is (3,-4)

So, the ,minimum value of f(x) is -4.

In g(x)= 2x

^{2}+ 8x + 3, a= 2, b= 8, and c=3. So, the equation for the axis of symmetry is given by

x = −(8)/2(2)

x = -8/4

x = -2

The equation of the axis of symmetry for g(x)= 2x^{2} + 8x + 3 is x = -2.

The x coordinate of the vertex is the same:

h =-2

The y coordinate of the vertex is :

k = f(h)

k = 2h^{2} + 8h + 3

k = 2(-2)^{2} + 8(-2) + 3

k = 8 - 16 + 3

k = -5

Therefore, the vertex is (-2 , -5)

The minimum value of g(x) will be the y-coordinate of vertex = -5

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -2x^{2} + 16x + 40

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

^{2}+ 16x + 40, a= -2, b= 16, and c= 40. So, the equation for the axis of symmetry is given by

x = −(16)/2(-2)

x = -16/-4

x = 4

The equation of the axis of symmetry for f(x)= -2x^{2} + 16x + 40 is x = 4.

The x coordinate of the vertex is the same:

h = 4

The y coordinate of the vertex is :

k = f(h)

k = -2h^{2} + 16h + 40

k = -2(4)^{2} + 16(4) + 40

k = -32 + 64 + 40

k = 72

Therefore, the vertex is (4 , 72)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -2(0)^{2} + 16(0) + 40

y = 0 + 0 + 40

y = 40

Therefore, Axis of symmetry is x = 4

Vertex is (4 , 72)

Y- intercept is 40.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -2x^{2} + 16x + 40

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

^{2}+ 16x + 40, a= -2, b= 16, and c= 40. So, the equation for the axis of symmetry is given by

x = −(16)/2(-2)

x = -16/-4

x = 4

The equation of the axis of symmetry for f(x)= -2x^{2} + 16x + 40 is x = 4.

The x coordinate of the vertex is the same:

h = 4

The y coordinate of the vertex is :

k = f(h)

k = -2h^{2} + 16h + 40

k = -2(4)^{2} + 16(4) + 40

k = -32 + 64 + 40

k = 72

Therefore, the vertex is (4 , 72)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -2(0)^{2} + 16(0) + 40

y = 0 + 0 + 40

y = 40

Therefore, Axis of symmetry is x = 4

Vertex is (4 , 72)

Y- intercept is 40.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = x^{2} - 6x + 12

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= x

^{2}- 6x + 12, a= 1, b= -6, and c= 12. So, the equation for the axis of symmetry is given by

x = −(-6)/2(1)

x = 6/2

x = 3

The equation of the axis of symmetry for f(x)= x^{2} - 6x + 12 is x = 3.

The x coordinate of the vertex is the same:

h = 3

The y coordinate of the vertex is :

k = f(h)

k = h^{2} – 6h + 12

k = (3)^{2} - 6(3) + 12

k = 9 – 18 + 12

k = 3

Therefore, the vertex is (3 , 3)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = (0)^{2} - 6(0) + 12

y = 0 + 0 + 12

y = 12

Therefore, Axis of symmetry is x = 3

Vertex is (3 , 3)

Y- intercept is 12.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = x^{2} - 6x + 12

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= x

^{2}- 6x + 12, a= 1, b= -6, and c= 12. So, the equation for the axis of symmetry is given by

x = −(-6)/2(1)

x = 6/2

x = 3

The equation of the axis of symmetry for f(x)= x^{2} - 6x + 12 is x = 3.

The x coordinate of the vertex is the same:

h = 3

The y coordinate of the vertex is :

k = f(h)

k = h^{2} – 6h + 12

k = (3)^{2} - 6(3) + 12

k = 9 – 18 + 12

k = 3

Therefore, the vertex is (3 , 3)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = (0)^{2} - 6(0) + 12

y = 0 + 0 + 12

y = 12

Therefore, Axis of symmetry is x = 3

Vertex is (3 , 3)

Y- intercept is 12.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 4x^{2} + 12x + 5

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= = 4x

^{2}+ 12x + 5, a= 4, b= 12, and c= 5. So, the equation for the axis of symmetry is given by

x = −(12)/2(4)

x = -12/8

x = -3/2 = -1.5

The equation of the axis of symmetry for f(x)= = 4x^{2} + 12x + 5 is x = -1.5.

The x coordinate of the vertex is the same:

h = -1.5

The y coordinate of the vertex is :

k = f(h)

k = 4h^{2} + 12h + 5

k = 4(-1.5)^{2} + 12(-1.5) + 5

k = 9 – 18 + 5

k = -4

Therefore, the vertex is (-1.5 , -4)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 4(0)^{2} + 12(0) + 5

y = 0 + 0 + 5

y = 5

Therefore, Axis of symmetry is x = -1.5

Vertex is ( -1.5 , -4)

Y- intercept is 5.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 4x^{2} + 12x + 5

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= = 4x

^{2}+ 12x + 5, a= 4, b= 12, and c= 5. So, the equation for the axis of symmetry is given by

x = −(12)/2(4)

x = -12/8

x = -3/2 = -1.5

The equation of the axis of symmetry for f(x)= = 4x^{2} + 12x + 5 is x = -1.5.

The x coordinate of the vertex is the same:

h = -1.5

The y coordinate of the vertex is :

k = f(h)

k = 4h^{2} + 12h + 5

k = 4(-1.5)^{2} + 12(-1.5) + 5

k = 9 – 18 + 5

k = -4

Therefore, the vertex is (-1.5 , -4)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 4(0)^{2} + 12(0) + 5

y = 0 + 0 + 5

y = 5

Therefore, Axis of symmetry is x = -1.5

Vertex is ( -1.5 , -4)

Y- intercept is 5.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 5x^{2} + 5x + 12

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 5x

^{2}+ 5x + 12 , a= 5, b= 5, and c= 12. So, the equation for the axis of symmetry is given by

x = −(5)/2(5)

x = -5/10

x = -1/2 = -0.5

The equation of the axis of symmetry for f(x)= 5x^{2} + 5x + 12 is x = -0.5.

The x coordinate of the vertex is the same:

h = -0.5

The y coordinate of the vertex is :

k = f(h)

k = 5(h)^{2} + 5h + 12

k = 5(-0.5)^{2} + 5(-0.5) + 12

k = 1.25 – 2.5 + 12

k = 10.75

Therefore, the vertex is (-0.5 , 10.75)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 5(0)^{2} + 5(0) + 12

y = 0 + 0 + 12

y = 12

Therefore, Axis of symmetry is x = -0.5

Vertex is ( -0.5 , 10.75)

Y- intercept is 12.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 5x^{2} + 5x + 12

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 5x

^{2}+ 5x + 12 , a= 5, b= 5, and c= 12. So, the equation for the axis of symmetry is given by

x = −(5)/2(5)

x = -5/10

x = -1/2 = -0.5

The equation of the axis of symmetry for f(x)= 5x^{2} + 5x + 12 is x = -0.5.

The x coordinate of the vertex is the same:

h = -0.5

The y coordinate of the vertex is :

k = f(h)

k = 5(h)^{2} + 5h + 12

k = 5(-0.5)^{2} + 5(-0.5) + 12

k = 1.25 – 2.5 + 12

k = 10.75

Therefore, the vertex is (-0.5 , 10.75)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 5(0)^{2} + 5(0) + 12

y = 0 + 0 + 12

y = 12

Therefore, Axis of symmetry is x = -0.5

Vertex is ( -0.5 , 10.75)

Y- intercept is 12.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -x^{2} - 2x – 5

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -x

^{2}- 2x – 5, a= -1, b= -2, and c= -5. So, the equation for the axis of symmetry is given by

x = −(-2)/2(-1)

x = 2/-2

x = -1

The equation of the axis of symmetry for f(x)= -x^{2} - 2x – 5 is x = -1.

The x coordinate of the vertex is the same:

h = -1

The y coordinate of the vertex is :

k = f(h)

k = -(h)^{2} - 2(h) - 5

k = -(-1)^{2} - 2(-1) - 5

k = -1 + 2 - 5

k = -4

Therefore, the vertex is (-1 , -4)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -(0)^{2} - 2(0) - 5

y = 0 - 0 - 5

y = -5

Therefore, Axis of symmetry is x = -1

Vertex is ( -1 , -4)

Y- intercept is -5.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -x^{2} - 2x – 5

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -x

^{2}- 2x – 5, a= -1, b= -2, and c= -5. So, the equation for the axis of symmetry is given by

x = −(-2)/2(-1)

x = 2/-2

x = -1

The equation of the axis of symmetry for f(x)= -x^{2} - 2x – 5 is x = -1.

The x coordinate of the vertex is the same:

h = -1

The y coordinate of the vertex is :

k = f(h)

k = -(h)^{2} - 2(h) - 5

k = -(-1)^{2} - 2(-1) - 5

k = -1 + 2 - 5

k = -4

Therefore, the vertex is (-1 , -4)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -(0)^{2} - 2(0) - 5

y = 0 - 0 - 5

y = -5

Therefore, Axis of symmetry is x = -1

Vertex is ( -1 , -4)

Y- intercept is -5.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 0.4x^{2} + 1.6x

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 0.4x

^{2}+ 1.6x, a= 0.4, b= 1.6, and c= 0. So, the equation for the axis of symmetry is given by

x = −(1.6)/2(0.4)

x = -1.6/0.8

x = -2

The equation of the axis of symmetry for f(x)= 0.4x^{2} + 1.6x is x = -2.

The x coordinate of the vertex is the same:

h = -2

The y coordinate of the vertex is :

k = f(h)

k = 0.4(h)^{2} + 1.6h

k = 0.4(-2)^{2} + 1.6(-2)

k = 1.6 – 3.2

k = -1.6

Therefore, the vertex is (-2 , -1.6)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 0.4(0)^{2} + 1.6(0)

y = 0 + 0

y = 0

Therefore, Axis of symmetry is x = -2

Vertex is ( -2 , -1.6)

Y- intercept is 0.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 0.4x^{2} + 1.6x

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 0.4x

^{2}+ 1.6x, a= 0.4, b= 1.6, and c= 0. So, the equation for the axis of symmetry is given by

x = −(1.6)/2(0.4)

x = -1.6/0.8

x = -2

The equation of the axis of symmetry for f(x)= 0.4x^{2} + 1.6x is x = -2.

The x coordinate of the vertex is the same:

h = -2

The y coordinate of the vertex is :

k = f(h)

k = 0.4(h)^{2} + 1.6h

k = 0.4(-2)^{2} + 1.6(-2)

k = 1.6 – 3.2

k = -1.6

Therefore, the vertex is (-2 , -1.6)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 0.4(0)^{2} + 1.6(0)

y = 0 + 0

y = 0

Therefore, Axis of symmetry is x = -2

Vertex is ( -2 , -1.6)

Y- intercept is 0.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -2x^{2} + 4x - 3

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

^{2}+ 4x -3, a= -2, b= 4, and c= -3. So, the equation for the axis of symmetry is given by

x = −(4)/2(-2)

x = -4/-4

x = 1

The equation of the axis of symmetry for f(x)= 2x^{2} + 4x - 3 is x = 1.

The x coordinate of the vertex is the same:

h = 1

The y coordinate of the vertex is :

k = f(h)

k = -2(h)^{2} + 4(h) - 3

k = -2(1)^{2} + 4(1) - 3

k = -2 + 4 - 3

k = -1

Therefore, the vertex is (1 , -1)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -2(0)^{2} + 4(0) - 3

y = 0 + 0 - 3

y = -3

Therefore, Axis of symmetry is x = 1

Vertex is ( 1 , -1)

Y- intercept is -3.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = -2x^{2} + 4x - 3

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= -2x

^{2}+ 4x -3, a= -2, b= 4, and c= -3. So, the equation for the axis of symmetry is given by

x = −(4)/2(-2)

x = -4/-4

x = 1

The equation of the axis of symmetry for f(x)= 2x^{2} + 4x - 3 is x = 1.

The x coordinate of the vertex is the same:

h = 1

The y coordinate of the vertex is :

k = f(h)

k = -2(h)^{2} + 4(h) - 3

k = -2(1)^{2} + 4(1) - 3

k = -2 + 4 - 3

k = -1

Therefore, the vertex is (1 , -1)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -2(0)^{2} + 4(0) - 3

y = 0 + 0 - 3

y = -3

Therefore, Axis of symmetry is x = 1

Vertex is ( 1 , -1)

Y- intercept is -3.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 2x^{2} + 8x + 2

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 2x

^{2}+ 8x + 2, a= 2, b= 8, and c= 2. So, the equation for the axis of symmetry is given by

x = −(8)/2(2)

x = -8/4

x = -2

The equation of the axis of symmetry for f(x)= 2x^{2} + 8x + 2 is x = -2.

The x coordinate of the vertex is the same:

h = -2

The y coordinate of the vertex is :

k = f(h)

k = 2(h)^{2} + 8(h) + 2

k = 2(-2)^{2} + 8(-2) + 2

k = 8 – 16 + 2

k = -6

Therefore, the vertex is (-2 , -6)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 2(0)^{2} + 8(0) + 2

y = 0 + 0 + 2

y = 2

Therefore, Axis of symmetry is x = -2

Vertex is ( -2 , -6)

Y- intercept is 2.

### Find the axis of symmetry, vertex and y-intercept of the function

f(x) = 2x^{2} + 8x + 2

^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 2x

^{2}+ 8x + 2, a= 2, b= 8, and c= 2. So, the equation for the axis of symmetry is given by

x = −(8)/2(2)

x = -8/4

x = -2

The equation of the axis of symmetry for f(x)= 2x^{2} + 8x + 2 is x = -2.

The x coordinate of the vertex is the same:

h = -2

The y coordinate of the vertex is :

k = f(h)

k = 2(h)^{2} + 8(h) + 2

k = 2(-2)^{2} + 8(-2) + 2

k = 8 – 16 + 2

k = -6

Therefore, the vertex is (-2 , -6)

For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = 2(0)^{2} + 8(0) + 2

y = 0 + 0 + 2

y = 2

Therefore, Axis of symmetry is x = -2

Vertex is ( -2 , -6)

Y- intercept is 2.

### Find the y-intercept of the following function

f(x) = -0.5x^{2} + x + 2

f(x) = -0.5x

^{2}+ x + 2

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = -0.5(0)

^{2 }+ (0) + 2

y = 2

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 2.

### Find the y-intercept of the following function

f(x) = -0.5x^{2} + x + 2

f(x) = -0.5x

^{2}+ x + 2

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = -0.5(0)

^{2 }+ (0) + 2

y = 2

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 2.

### Find the y-intercept of the following function

f(x) = - x^{2} – 2x + 3

f(x) = x

^{2}– 2x + 3

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = (0)

^{2 }– 2(0) + 3

y = 3

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 3.

### Find the y-intercept of the following function

f(x) = - x^{2} – 2x + 3

f(x) = x

^{2}– 2x + 3

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = (0)

^{2 }– 2(0) + 3

y = 3

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 3.

### Find the y-intercept of the following function

f(x) = 3x^{2} + 6x + 5

f(x) = 3x

^{2}+ 6x + 5

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 3(0)

^{2 }+ 6(0) + 5

y = 5

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 5.

### Find the y-intercept of the following function

f(x) = 3x^{2} + 6x + 5

f(x) = 3x

^{2}+ 6x + 5

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 3(0)

^{2 }+ 6(0) + 5

y = 5

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is 5.

### Find the y-intercept of the following function

f(x) = -2x^{2} – 8x – 7

f(x) = 2x

^{2}– 8x – 7

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 2(0)

^{2 }– 8(0) – 7

y = -7

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is -7.

### Find the y-intercept of the following function

f(x) = -2x^{2} – 8x – 7

f(x) = 2x

^{2}– 8x – 7

We will compare the given equation with the standard equation f(x)=ax

^{2}+bx+c.

We know that for y intercept , x = 0

So, for finding y- intercept

f(x) = y = 2(0)

^{2 }– 8(0) – 7

y = -7

On comparing with the standard form y-intercept is equal to c

y-intercept of given quadratic function is -7.