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

Question

1633 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-10T20:49:52

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

$y=(x-4)^2+1$

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

$(x-4)^2=(x^2)+(2*x*-4)+(-4)^2$ =

$x^2-8x+16$

Substitute the equation back to the original equation.

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

$y=x^2-8+17$

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