For $f(x+1)$, think about how the $x$-coordinate of a point changes so that the input to $f$ stays the same.

After handling the horizontal shift, apply the outside operations to the old $y$-value in the order: multiply by $-2$, then add $3$.

Using the x-intercepts $-1$ and $5$ and the y-intercept $5$, type

$y=a(x+1)(x-5)$

and adjust the slider $a$ until the graph matches the given parabola (it will match when $a=-1$).

In a new line, type

$y=-2\left(a((x+1)+1)((x+1)-5)\right)+3$

(or, if you set $a=-1$, you can type $y=-2(- (x+2)(x-4))+3$).

Type each answer choice as a point, like $P=(1,-15)$, and see which point lies on the transformed graph.

From the graph of $y=f(x)$, the vertex is at $(2,9)$, so $f(2)=9$.

In $y=-2f(x+1)+3$, the input to $f$ is $x+1$.

To use the point where the input equals 2, set $x+1=2$, so $x=1$.

Now compute the new $y$-value at $x=1$:

So the point on the transformed graph is $(1,-15)$.