Maths-
General
Easy

Question

Find the axis of symmetry, vertex and y-intercept of the function
f(x) = -x2 - 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)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= -x2 - 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)= -x2 - 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.

    Related Questions to study

    General
    Maths-

    Find the axis of symmetry, vertex and y-intercept of the function
    f(x) = 0.4x2 + 1.6x

    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 0.4x2 + 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)= 0.4x2 + 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) = 0.4x2 + 1.6x

    Maths-General
    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 0.4x2 + 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)= 0.4x2 + 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.

    General
    Maths-

    Find the axis of symmetry, vertex and y-intercept of the function
    f(x) = -2x2 + 4x - 3

    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= -2x2 + 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)= 2x2 + 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) = -2x2 + 4x - 3

    Maths-General
    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= -2x2 + 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)= 2x2 + 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.

    General
    Maths-

    Find the axis of symmetry, vertex and y-intercept of the function
    f(x) = 2x2 + 8x + 2

    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 2x2 + 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)= 2x2 + 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) = 2x2 + 8x + 2

    Maths-General
    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 2x2 + 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)= 2x2 + 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.

    parallel
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = -0.5x2 + x + 2

    We have given a function
    f(x) = -0.5x2 + x + 2
    We will compare the given equation with the standard equation f(x)=ax2+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) = -0.5x2 + x + 2

    Maths-General
    We have given a function
    f(x) = -0.5x2 + x + 2
    We will compare the given equation with the standard equation f(x)=ax2+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.
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = - x2 – 2x + 3

    We have given a function
    f(x) = x2 – 2x + 3
    We will compare the given equation with the standard equation f(x)=ax2+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) = - x2 – 2x + 3

    Maths-General
    We have given a function
    f(x) = x2 – 2x + 3
    We will compare the given equation with the standard equation f(x)=ax2+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.
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = 3x2 + 6x + 5

    We have given a function
    f(x) = 3x2 + 6x + 5
    We will compare the given equation with the standard equation f(x)=ax2+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) = 3x2 + 6x + 5

    Maths-General
    We have given a function
    f(x) = 3x2 + 6x + 5
    We will compare the given equation with the standard equation f(x)=ax2+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.
    parallel
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = -2x2 – 8x – 7

     We have given a function
    f(x) = 2x2 – 8x – 7
    We will compare the given equation with the standard equation f(x)=ax2+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) = -2x2 – 8x – 7

    Maths-General
     We have given a function
    f(x) = 2x2 – 8x – 7
    We will compare the given equation with the standard equation f(x)=ax2+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.
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = 0.3x2 + 0.6x – 0.7

    We have given a function
    f(x) = 0.3x2 + 0.6x – 0.7
    We will compare the given equation with the standard equation f(x)=ax2+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) = 0.3x2 + 0.6x – 0.7

    Maths-General
    We have given a function
    f(x) = 0.3x2 + 0.6x – 0.7
    We will compare the given equation with the standard equation f(x)=ax2+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.
    General
    Maths-

    Find the y-intercept of the following function
    f(x) = 2x2 – 4x – 6

    We have given a function
    f(x) = 2x2 – 4x – 6
    We will compare the given equation with the standard equation f(x)=ax2+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) = 2x2 – 4x – 6

    Maths-General
    We have given a function
    f(x) = 2x2 – 4x – 6
    We will compare the given equation with the standard equation f(x)=ax2+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.
    parallel
    General
    Maths-

    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

    Maths-General

    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
    General
    Maths-

    Estimate the coordinates of the vertex of the graph of f(x) = 1.25x2 -2x -1 below. Then explain how to find the exact coordinates

    We have given a function
    f(x) = 1.25x2 -2x -1
    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 1.25x2 -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)= 1.25x2 -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 = 1.25h2 -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) = 1.25x2 -2x -1 below. Then explain how to find the exact coordinates

    Maths-General
    We have given a function
    f(x) = 1.25x2 -2x -1
    This quadratic function is in standard form, f(x)=ax2+bx+c.
    For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
    In f(x)= 1.25x2 -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)= 1.25x2 -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 = 1.25h2 -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)
    General
    Maths-

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

    For identifying the y- intercept we have to put the value of x as 0 in the given function so that we can obtain the y-intercept
    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

    Maths-General
    For identifying the y- intercept we have to put the value of x as 0 in the given function so that we can obtain the y-intercept
    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.
    parallel
    General
    Maths-

    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

    We have given a function of water balloon tossed into air,
    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

    Maths-General
    We have given a function of water balloon tossed into air,
    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.
    General
    Maths-

    Sage began graphing f(x) = -2x2 + 4x + 9 by finding the axis of symmetry x = -1. Explain the error Sage made?

    We have given the equation f(x) = -2x2 + 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) = -2x2 + 4x + 9 by finding the axis of symmetry x = -1. Explain the error Sage made?

    Maths-General
    We have given the equation f(x) = -2x2 + 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

    General
    Maths-

    How are the form and graph of f(x) = ax2 + bx + c similar to the form and graph of g(x) = ax2 + bx? How are they different?

    We have given the two functions
    f(x) = ax2 + bx + c
    g(x) = ax2 + 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) = ax2 + bx + c similar to the form and graph of g(x) = ax2 + bx? How are they different?

    Maths-General
    We have given the two functions
    f(x) = ax2 + bx + c
    g(x) = ax2 + 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)
    parallel

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