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

Answer:

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

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

Step-by-step 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