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dusya [7]
3 years ago
6

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

Mathematics
1 answer:
Likurg_2 [28]3 years ago
8 0

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)

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