# How do you solve y=-(x-4)^2+1?

Jul 10, 2015

The solution is $y = {x}^{2} - 8 + 17$.

#### Explanation:

$y = {\left(x - 4\right)}^{2} + 1$

${\left(x - 4\right)}^{2}$ represents a square of a difference, the formula of which is ${\left(x - 4\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$, where $a = x ,$ and $b = - 4$.

${\left(x - 4\right)}^{2} = \left({x}^{2}\right) + \left(2 \cdot x \cdot - 4\right) + {\left(- 4\right)}^{2}$ =

${x}^{2} - 8 x + 16$

Substitute the equation back to the original equation.

$y = {x}^{2} - 8 x + 16 + 1$=

$y = {x}^{2} - 8 + 17$