Answer:
Option C. x = -2,6 extreme value at (2.16)
Step-by-step explanation:
we have
[tex]y=-x^2+4x+12[/tex]
This is the equation of a vertical parabola open down
The vertex is a maximum (extreme value)
Convert the equation into vertex form
[tex]y=-x^2+4x+12[/tex]
Complete the square
Group terms that contain the same variable and move the constant term to the left side
[tex]y-12=-x^2+4x[/tex]
Factor -1
[tex]y-12=-(x^2-4x)[/tex]
Remember to balance the equation by adding the same constants to each side.
[tex]y-12-4=-(x^2-4x+4)[/tex]
Rewrite as perfect squares
[tex]y-16=-(x-2)^2[/tex]
[tex]y=-(x-2)^2+16[/tex] -----> equation of the parabola in vertex form
The vertex is the point (2,16) ----> is a maximum (extreme value)
Determine the solutions of the quadratic equation
For y=0
[tex]0=-(x-2)^2+16[/tex]
[tex](x-2)^2=16[/tex]
square root both sides
[tex](x-2)=(+/-)4[/tex]
[tex]x=2(+/-)4[/tex]
[tex]x=2(+)4=6[/tex]
[tex]x=2(-)4=-2[/tex]
therefore
The solutions are x=-2 and x=6
The extreme value is (2,16)
