Question

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

Answer 1

Step-by-step explanation:

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

`y=ax^2+bx+c`

is:

`x_v=(-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=(-(-12))/(2\cdot4)=(12)/(8)=(3)/(2)`

And the y-value:

`\begin{gathered} y_v=4x^2_v-12x_v+5 \\ y_v=4\cdot(3^2)/(2^2)-12\cdot(3)/(2)+5 \\ y_v=-4 \end{gathered}`

Answer:

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