Answer
$1$ and $5$.
Work Step by Step
The given function is
$f(x)=-2(x-3)^2+8$
Plug $f(x)=0$ into the given function.
$\Rightarrow 0=-2(x-3)^2+8$
Factor out $2$.
$\Rightarrow 0=2[-(x-3)^2+4]$
Divide both sides by $2$.
$\Rightarrow \frac{0}{2}=\frac{2[-(x-3)^2+4]}{2}$
Simplify.
$\Rightarrow 0=-(x-3)^2+4$
Use the special formula $(A-B)^2=A^2-2AB+B^2$
We have $A=x$ and $B=3$
$\Rightarrow 0=-[x^2-2(x)(3)+3^2]+4$
Simplify.
$\Rightarrow 0=-[x^2-6x+9]+4$
Use the distributive property.
$\Rightarrow 0=-x^2+6x-9+4$
Simplify.
$\Rightarrow 0=-x^2+6x-5$
Rewrite the middle term $6x$ as $5x+1x$.
$\Rightarrow 0=-x^2+5x+1x-5$
Group the terms.
$\Rightarrow 0=(-x^2+5x)+(1x-5)$
Factor groups.
$\Rightarrow 0=-x(x-5)+1(x-5)$
Factor out $(x-5)$.
$\Rightarrow 0=(x-5)(-x+1)$
Set each factor equal to zero.
$\Rightarrow x-5=0$ or $-x+1=0$
Isolate $x$.
$\Rightarrow x=5$ or $x=1$
Hence, the $x-$ intercepts are $1$ and $5$.
