Question

Rewrite the function by completing the square.
f(x) = x^2 – 10x—96

Answer 1

(x - 5)² -121 = 0

Explanation:

if you need to find the roots you can take the square root of each side:

(x-5)² = 121

(x-5)² = 121

square root of (x-5)² is x-5

square root of 121 is ±11

first root: x-5 = 11

x = 16

second root: x-5 = -11

x = -6

Answer 2

`f(x) = (x - 5)^(2) - 121`.

Explanation:

The goal is to rewrite
`f(x)` in the vertex form
`a\, (x - h)^(2) + k` by completing the square (where
`a`,
`h`, and
`k` are constants.)

Expand the vertex form expression:

`\begin{aligned}& a\, (x - h)^(2) + k\\ &= a\, (x - h)\, (x - h) + k \\ &= a\, \left(x^2 - h\, x - h\, x + h^2\right) + k \\ &= a\, \left(x^2 - 2\, h\, x + h^2\right) + k\\ &= a\, x^2 - 2\, a\, h\, x + \left(a\, h^2 + k\right) \end{aligned}`.

Compare this expression to
`f(x) = x^2 - 10\, x - 96` and solve for the constants
`a`,
`h`, and
`k`. Make sure that the coefficient of each term matches:

• Coefficient for the
  `x^2` term:
  `a` in the expanded expression and
  `1` in the expression for
  `f(x)`. Hence,
  `a = 1`.

• Coefficient for the
  `x` term:
  `(-2\, a\, h)` in the expanded expression and
  `(-10)` in the expression for
  `f(x)`. Hence,
  `-2\, a\, h = -10`.

• Coefficient for the constant term:
  `\left(a\, h^2 + k\right)` in the expanded expression and
  `(-96)` in the expression for
  `f(x)`. Hence,
  `a\, h^(2) + k = -96`.

Substitute
`a = 1` into the second equation,
`-2\, a\, h = -10`, and solve for
`h`.

`-2 \, h = -10`.

`h = 5`.

Substitute both
`a = 1` and
`h = 5` into the third equation,
`a\, h^(2) + k = -96`, and solve for
`k`.

`5^2 + k = -96`.

`k = -121`.

Therefore,
`a\, (x - h)^(2) + k` becomes
`(x - 5)^2 + (-121)`.

Hence, the vertex form of the parabola
`f(x)` would be:

`f(x) = (x - 5)^(2) - 121`.