Question

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

Answer 1

Answer:

(x - 5)² -121 = 0

Step-by-step 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

Answer:

$f(x) = (x - 5)^{2} - 121$.

Step-by-step 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:

$f(x) = (x - 5)^{2} - 121$.