Maths-
General
Easy

Question

Find the axis of symmetry, vertex and y-intercept of the function
f(x) = x2 - 6x + 12

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: 12


    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 - 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)= x2 - 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 = h2 – 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.

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