Question

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

f(x) = -x^{2} - 2x – 5

Hint:

### For a quadratic function is in standard form, f(x)=ax2+bx+c.

A vertical line passing through the vertex is called the axis of symmetry for the parabola.

Axis of symmetry x=−b/2a

Vertex The vertex of the parabola is located at a pair of coordinates which we will call (*h, k*). where h is value of x in axis of symmetry formula and k is f(h).

The *y*-intercept is the point where a graph crosses the *y*-axis. In other words, it is the value of *y* when x=0.

## The correct answer is: 0

### This quadratic function is in standard form, f(x)=ax^{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.

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:

The y coordinate of the vertex is :

^{2}- 2(h) - 5

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

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

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

^{2}- 2(0) - 5

Therefore, Axis of symmetry is x = -1

Vertex is ( -1 , -4)

Y- intercept is -5.

### Related Questions to study

### 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.

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

f(x) = 0.3x^{2} + 0.6x – 0.7

f(x) = 0.3x

^{2}+ 0.6x – 0.7

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 = 0.3(0)

^{2}+ 0.6(0) – 0.7

y = -0.7

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

y-intercept of given quadratic function is -0.7.

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

f(x) = 0.3x^{2} + 0.6x – 0.7

f(x) = 0.3x

^{2}+ 0.6x – 0.7

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 = 0.3(0)

^{2}+ 0.6(0) – 0.7

y = -0.7

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

y-intercept of given quadratic function is -0.7.

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

f(x) = 2x^{2} – 4x – 6

f(x) = 2x

^{2}– 4x – 6

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 }– 4(0) – 6

y = -6

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

y-intercept of given quadratic function is -6.

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

f(x) = 2x^{2} – 4x – 6

f(x) = 2x

^{2}– 4x – 6

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 }– 4(0) – 6

y = -6

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

y-intercept of given quadratic function is -6.

### Points (2, -1), (-2, 7), (1, -2), (0, -1) and (4, 7) lie on graph of a quadratic function

1.Find axis of symmetry of graph

2.Find the vertex

3.Find the y-intercept

4.Over what interval does the function increase

We can see that the axis of symmetry passes through x= 1

Therefore, the vertex point is (1,-2)

Y intercept is the point at which x = 0 which is (0,-1)

y- intercept= -1

### Points (2, -1), (-2, 7), (1, -2), (0, -1) and (4, 7) lie on graph of a quadratic function

1.Find axis of symmetry of graph

2.Find the vertex

3.Find the y-intercept

4.Over what interval does the function increase

We can see that the axis of symmetry passes through x= 1

Therefore, the vertex point is (1,-2)

Y intercept is the point at which x = 0 which is (0,-1)

y- intercept= -1

### Estimate the coordinates of the vertex of the graph of f(x) = 1.25x^{2} -2x -1 below. Then explain how to find the exact coordinates

f(x) = 1.25x

^{2}-2x -1

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

^{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)= 1.25x

^{2}-2x -1, a= 1.25, b= -2, and c= -1. So, the equation for the axis of symmetry is given by

x = −(-2)/2(1.25)

x = 2/2.5

x = 0.8

The equation of the axis of symmetry for f(x)= 1.25x

^{2}-2x -1 is x = 0.8.

The x coordinate of the vertex is the same:

h = 0.8

The y coordinate of the vertex is :

k = f(h)

k = 1.25h

^{2}-2h -1

k = 1.25(0.8)

^{2}- 2(0.8) - 1

k = 0.8 - 1.6 – 1

k = -1.8

Therefore, the vertex is (0.8 , -1.8)

### Estimate the coordinates of the vertex of the graph of f(x) = 1.25x^{2} -2x -1 below. Then explain how to find the exact coordinates

f(x) = 1.25x

^{2}-2x -1

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

^{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)= 1.25x

^{2}-2x -1, a= 1.25, b= -2, and c= -1. So, the equation for the axis of symmetry is given by

x = −(-2)/2(1.25)

x = 2/2.5

x = 0.8

The equation of the axis of symmetry for f(x)= 1.25x

^{2}-2x -1 is x = 0.8.

The x coordinate of the vertex is the same:

h = 0.8

The y coordinate of the vertex is :

k = f(h)

k = 1.25h

^{2}-2h -1

k = 1.25(0.8)

^{2}- 2(0.8) - 1

k = 0.8 - 1.6 – 1

k = -1.8

Therefore, the vertex is (0.8 , -1.8)

### To identify the y-intercept of quadratic function, would you choose to use vertex form or standard form? Explain

If we see the standard form, we get an constant number c on putting the value of x=0. This value of c is called as y-intercept.

But in vertex form it is not so easy to get y-intercept because in this form the constant term is not distinct as in the standard form

So, Standard form is the right choice for identifying the y-intercept of quadratic function.

### To identify the y-intercept of quadratic function, would you choose to use vertex form or standard form? Explain

If we see the standard form, we get an constant number c on putting the value of x=0. This value of c is called as y-intercept.

But in vertex form it is not so easy to get y-intercept because in this form the constant term is not distinct as in the standard form

So, Standard form is the right choice for identifying the y-intercept of quadratic function.

### A water balloon is tossed into the air. The function h(x) = -0.5(x-4)^{2} + 9 gives the height, in feet, of the balloon from the surface of a pool as a function of the balloon’s horizontal distance from where it was first tossed. Will the balloon hit the ceiling 12 ft above the pool? Explain

h(x) = -0.5(x-4)

^{2}+ 9

The given equation is in the vertex form

We know vertex is the point at maximum height .

And y coordinate of vertex is the maximum height of the ball

On comparing with the vertex form we get that

h = 4

k = 9

Therefore, here vertex is (4, 9)

Here, we have y-coordinate as 9

So, the maximum height of ball is 9

We have given the ceiling height as 12 ft

So the height of ceiling is more than maximum height of ball , so ball will not hit the ceiling.

### A water balloon is tossed into the air. The function h(x) = -0.5(x-4)^{2} + 9 gives the height, in feet, of the balloon from the surface of a pool as a function of the balloon’s horizontal distance from where it was first tossed. Will the balloon hit the ceiling 12 ft above the pool? Explain

h(x) = -0.5(x-4)

^{2}+ 9

The given equation is in the vertex form

We know vertex is the point at maximum height .

And y coordinate of vertex is the maximum height of the ball

On comparing with the vertex form we get that

h = 4

k = 9

Therefore, here vertex is (4, 9)

Here, we have y-coordinate as 9

So, the maximum height of ball is 9

We have given the ceiling height as 12 ft

So the height of ceiling is more than maximum height of ball , so ball will not hit the ceiling.

### Sage began graphing f(x) = -2x^{2} + 4x + 9 by finding the axis of symmetry x = -1. Explain the error Sage made?

^{2}+ 4x + 9

We will first find the axis of symmetry ‘

x = -b/ 2a

x = -4 / 2 (-2)

x = -4/ -4

x = 1

This detects the error made by sage that he finds axis of symmetry as x = -1

### Sage began graphing f(x) = -2x^{2} + 4x + 9 by finding the axis of symmetry x = -1. Explain the error Sage made?

^{2}+ 4x + 9

We will first find the axis of symmetry ‘

x = -b/ 2a

x = -4 / 2 (-2)

x = -4/ -4

x = 1

This detects the error made by sage that he finds axis of symmetry as x = -1

### How are the form and graph of f(x) = ax^{2} + bx + c similar to the form and graph of g(x) = ax^{2} + bx? How are they different?

f(x) = ax

^{2}+ bx + c

g(x) = ax

^{2}+ bx

By analyzing both the equations we get that both the equation are of same format

In first equation the constant c is present whereas in second equation constant c is zero. So we know that in the standard form this constant represents the y- intercept of the curve.

So the first curve intercepts y axis at (0,c)

And second curve intercepts y axis at (0,0)

### How are the form and graph of f(x) = ax^{2} + bx + c similar to the form and graph of g(x) = ax^{2} + bx? How are they different?

f(x) = ax

^{2}+ bx + c

g(x) = ax

^{2}+ bx

By analyzing both the equations we get that both the equation are of same format

In first equation the constant c is present whereas in second equation constant c is zero. So we know that in the standard form this constant represents the y- intercept of the curve.

So the first curve intercepts y axis at (0,c)

And second curve intercepts y axis at (0,0)