math help pls???

Question

write the function f(x)=x^2-2x-2 in the form f(x)=(x-h)^2+k

thanks!!

Answer 1

Hi,

f(x)=x^2 - 2x - 2

Multiply 1/2 times the number in front of the x term. Square the answer. Add it behind the x term and subtract it behind the constant, as below.

1/2 * -2 = -1

(-1)^2 = 1 Add and subtract this.

f(x)=x^2 - 2x +1 - 2 -1

Think of this as:

f(x)=(x^2 - 2x +1) - 2 -1

Factor inside the parentheses and combine like terms at the end.

f(x)=(x - 1)^2 - 3

This is in the form you wanted. h is 1 and k is -3 which means the vertex on your graph would be at (1,-3).

I hope that helps!! :-)

Answer 2

To complete a square with -2x as the second term, the square would be (x-1)^2. This provides a final constant of +1, thus k = -3.

Answer 3

f(x) = x^2 - 2x - 2

To put this in the form f(x) = (x - h)^2 + k, you had to complete the square.

First step: add and subtract "half squared" of the coefficient of x.

In this case, the coefficient of x is -2. Take half of this, which is -1, and then square it; (-1) squared is 1. Add and them subtract this.

f(x) = x^2 - 2x + {1} - 2 - {1}

I put { } brackets to show you what I added. Remember that doing this does not change the expression, because we added and subtracted 1; the same as adding 0.

The first three terms now form a perfect square trionomial; factor it as such.

f(x) = (x - 1)^2 - 2 - 1

f(x) = (x - 1)^2 - 3

Answer 4

complete the square:

x^2-2x +1 -3 (since 1 -3 = -2)

(x-1)^2 - 3

h =1 , k = -3

Answer 5

complete the square:
x ^2-2x-2=0
x ^2-2x+1=3
factor:
(x-1) (x-1)=3
(x-1) ^2=3

answer:
f(x)=(x-1) ^2-3

Answer 6

x^2 - 2x - 2 = x^2 - 2x + (1 - 1) - 2 = (x^2 - 2x + 1) - 1 -2 =
(x - 1)^2 - 3

Answer 7

assuming k and h are constants

f(x) = (x-1)^2 -3

i just took the -2x, and knowing that the only way to get it by squaring a binomial is using (x-1).