Question

Complete the square for the following quadratic equation to determine its solutions and the location of its extreme value

Answer 1

Answer:

Option C. x = -2,6  extreme value at (2.16)

Step-by-step explanation:

we have

$y=-x^2+4x+12$

This is the equation of a vertical parabola open down

The vertex is a maximum (extreme value)

Convert the equation into vertex form

$y=-x^2+4x+12$

Complete the square

Group terms that contain the same variable and move the constant term to the left side

$y-12=-x^2+4x$

Factor -1

$y-12=-(x^2-4x)$

Remember to balance the equation by adding the same constants to each side.

$y-12-4=-(x^2-4x+4)$

Rewrite as perfect squares

$y-16=-(x-2)^2$

$y=-(x-2)^2+16$ -----> 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

$0=-(x-2)^2+16$

$(x-2)^2=16$

square root both sides

$(x-2)=(+/-)4$

$x=2(+/-)4$

$x=2(+)4=6$

$x=2(-)4=-2$

therefore

The solutions are x=-2 and x=6

The extreme value is (2,16)