SL 2.11—Transformation of functions

The function is written in vertex form $y=p(x-h{)}^2+k$, where the vertex is at $(h,k)$. Here $f(x)=3(x+1{)}^2-5$.

Rewrite as $f(x)=3(x-(-1){)}^2+(-5)$. Comparing with $y=p(x-h{)}^2+k$, we get $h=-1$ and $k=-5$.

The expression $(x+1)$ means $x-(-1)$, so the horizontal shift is $-1$ (one unit left), not $+1$. The vertex is therefore at $(-1,-5)$.

The vertex is $(-1,-5)$.

We need the vertex of $f(x)=3(x+1{)}^2-5$. The vertex is the point where the squared term equals zero.

Setting $x=1$ gives $f(1)=3(2{)}^2-5=7eq-5$. The point $(1,-5)$ does not lie on the graph, so this is incorrect.

Setting $x=-1$ gives $f(-1)=3(0{)}^2-5=-5$, not $5$. The $y$-coordinate is wrong here.

The coefficient $3$ is the vertical stretch factor, not the $x$-coordinate of the vertex. This is a common misread of the vertex form.

Setting $x=-1$: $f(-1)=3(0{)}^2-5=-5$. The vertex is confirmed at $(-1,-5)$.

Begin with $y=x^2$.

A reflection in the $x$-axis replaces $y$ with $-y$, giving $y=-x^2$. Every $y$-value is negated.

Adding $b=4$ outside the function gives $y=-x^2+4$. Each point $(x,y)$ moves to $(x,y+4)$.

The resulting equation is $y=-x^2+4$.

We need to reflect $y=x^2$ in the $x$-axis (negate $y$-values) and then shift upward by 4.

This applies only the vertical translation without the reflection. The parabola still opens upward, which is incorrect.

This reflects in the $x$-axis and shifts the parabola 4 units to the left (horizontal), not upward. The translation direction is wrong.

This reflects in the $x$-axis and then shifts downward by 4 units ($b=-4$), which is opposite to the required upward translation.

This correctly applies both transformations: reflection in the $x$-axis (negative sign) and vertical translation 4 up ($+4$ outside). This is the correct answer.