Question

Consider h(x) = 3x^2 +2x - 8

Answer 1

Answer:

Vertex: $(-\frac{1}{3},-\frac{25}{3})$

Y-intercept: $(0,-8)$

Step-by-step explanation:

The x-coordinate of the vertex would be $x=\frac{-b}{2a}$ and the y-coordinate of the vertex is whatever the output is given the value of x.

Therefore, the x-coordinate of the vertex is $x=\frac{-b}{2a}=\frac{-2}{2(3)}=\frac{-2}{6}=-\frac{1}{3}$

This means the y-coordinate of the vertex is $y=3x^2+2x-8=3(-\frac{1}{3})^2+2(-\frac{1}{3})-8=3(\frac{1}{9})-\frac{2}{3}-8=\frac{1}{3}-\frac{2}{3}-8=-\frac{1}{3}-8=-\frac{1}{3}-\frac{24}{3}=-\frac{25}{3}$

So, the vertex is $(-\frac{1}{3},-\frac{25}{3})$

The y-intercept of a function is the y-value at which x=0, or the y-value when the function crosses the y-axis. Therefore, if we plug x=0 into the function, we see that $h(0)=3(0)^2+2(0)-8=0+0-8=-8$, so our y-intercept is -8 or (0,-8).

I attached a graph below to help you visualize the vertex and y-intercept given the function.