Find the $y$-intercept and any $x$-intercept(s) of the parabola with equation $y=-4(x-4{)}^2+16$

If the parabola doesn't have any $x$-intercepts, type DNE, meaning "does not exist."
If the parabola has two $x$-intercepts, use a comma to separate them.
$y$-intercept:
$x$-intercept(s):

Step 1 ：To find the $y$-intercept, set $x=0$ and solve for $y$ in the equation $y=-4(x-4{)}^2+16$

Step 2 ：Substitute $x=0$ into the equation to get $y=-4(0-4{)}^2+16$

Step 3 ：Simplify the equation to get $y=-4(-4{)}^2+16$

Step 4 ：Calculate the value to get $y=-4(16)+16$

Step 5 ：Simplify further to get $y=-64+16$

Step 6 ：Finally, calculate the $y$-intercept to get $y=-48$

Step 7 ：To find the $x$-intercept(s), set $y=0$ and solve for $x$ in the equation $y=-4(x-4{)}^2+16$

Step 8 ：Substitute $y=0$ into the equation to get $0=-4(x-4{)}^2+16$

Step 9 ：Rearrange the equation to get $4(x-4{)}^2=16$

Step 10 ：Divide both sides by 4 to get $(x-4{)}^2=4$

Step 11 ：Take the square root of both sides to get $x-4=\pm 2$

Step 12 ：Solve for $x$ to get the $x$-intercept(s) as $x=4\pm 2$

Step 13 ：Calculate the values to get the $x$-intercept(s) as $x=2$ and $x=6$

Step 14 ：The final answer for the $y$-intercept is $\boxed{{\displaystyle -48}}$

Step 15 ：The final answer for the $x$-intercept(s) is $\boxed{{\displaystyle 2,6}}$