Solve y=-(x-4)^2-3

$$y = - ( x - 4 ) ^ { 2 } - 3$$

Solve for y

$$y=-\left(\left(x-4\right)^{2}+3\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$y=-\left(x^{2}-8x+16\right)-3$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$y=-x^{2}+8x-16-3$$

Subtract 3 from -16 to get -19.

$$y=-x^{2}+8x-19$$

Solve for x (complex solution)

$$x=-i\sqrt{y+3}+4$$

$$x=i\sqrt{y+3}+4$$

Solve for x

$$x=-\sqrt{-y-3}+4$$

$$x=\sqrt{-y-3}+4\text{, }y\leq -3$$