Question

which statements are true about the graph of the function f(x) = 6x – 4 x2? check all that apply. the vertex form of the function is f(x) = (x – 2)2 2. the vertex of the function is (–3, –13). the axis of symmetry for the function is x = 3. the graph increases over the interval (–3, ). the function does not cross the x-axis.

Answer 1

A. The function in vertex form if f(x)=(x-4)^2-1
D. The y-intercept of the function is (0,5).
E. The function crosses the x-axis twice.

Answer 2

The vertex of the function is the point
`(-3,-13)`

The graph increase over the interval--------> (-3,∞)

Explanation:

we have

`f(x)=6x-4+x^(2)`

1) Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)+4=x^(2)+6x`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)+4+9=x^(2)+6x+9`

`f(x)+13=x^(2)+6x+9`

Rewrite as perfect squares

`f(x)+13=(x+3)^(2)`

`f(x)=(x+3)^(2)-13` -----> function in vertex form

2) Find the vertex

The vertex of the function is the point
`(-3,-13)`

3) Find the axis of symmetry

we know that

In a vertical parabola, the axis of symmetry is equal to the x-coordinate of the vertex

The x-coordinate of the vertex in this problem is equal to
`x=-3`

therefore

the equation of the axis of symmetry is
`x=-3`

4) Find the increase-decrease intervals

The graph increase over the interval--------> (-3,∞)

The graph decrease over the interval--------> (-∞,-3)

see the attached figure to better understand the problem

5) Find the x-intercepts of the function

we know that

the x-intercepts are the values of x when the value of the function is equal to zero

In this problem the x-intercepts are

`(-6.61,0)` and
`(0.61,0)`

so

The function cross the x-axis twice

see the attached figure