Question

PLEASEEEESSSSS HELPPPPPPPPP!!!!!NEEDDDDDDDD HELP URGENT IT'S A PRACTICE ASSESSMENT ITS NOT GRADED!!!!!!!!!!!!!!!!!!!

Answer 1

Explanation:

The equation for the x-value of the vertex of a quadratic equation:

$y=ax^2+bx+c$

is:

$x_v=\frac{-b}{2a}$

And to find the y-value of the vertex we have to replace its x-value into the quadratic equation.

In this problem we have b = -12 and a = 4

$x_v=\frac{-(-12)}{2\cdot4}=\frac{12}{8}=\frac{3}{2}$

And the y-value:

$\begin{gathered} y_v=4x^2_v-12x_v+5 \\ y_v=4\cdot\frac{3^2}{2^2}-12\cdot\frac{3}{2}+5 \\ y_v=-4 \end{gathered}$

Answer:

The vertex of this parabola is located at point (3/2, -4)