Maths-
General
Easy

Question

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

hintHint:

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: 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.

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