In $y = (x + 3)^2 - 5$, how could you rewrite $x + 3$ so that it looks like $x - h$? What does that tell you about the horizontal movement?

In vertex form $y = a(x - h)^2 + k$, the number $k$ moves the graph up or down. In $h(x)$, what role does the $-5$ play?

In Desmos, enter y = x^2 to see the original parabola with vertex at $(0,0)$.

On a new line, enter y = (x + 3)^2 - 5. Observe where the new parabola's vertex is located relative to the original vertex at $(0,0)$.

Note how many units left or right and how many units up or down you must move from $(0,0)$ to reach the new vertex. That horizontal and vertical movement tells you the graph's shift from $y = x^2$ to $y = h(x)$.

The standard "vertex form" of a quadratic is

This represents the parabola $y = x^2$ that has been:

• shifted $h$ units horizontally (right if $h>0$, left if $h<0$), and
• shifted $k$ units vertically (up if $k>0$, down if $k<0$).

The vertex of $y = x^2$ is at $(0,0)$, and the vertex of $y = a(x - h)^2 + k$ is at $(h,k)$.

We are given

Rewrite the inside part $x + 3$ in the form $x - h$:

• $x + 3$ is the same as $x - (-3)$.

So $h(x)$ can be viewed as

which matches $y = a(x - h)^2 + k$ with $a = 1$, $h = -3$, and $k = -5$.

From the matching in the previous step:

• $h = -3$ means the vertex moves from $(0,0)$ to $(-3,0)$ horizontally, which is a shift 3 units to the left.
• $k = -5$ means the vertex then moves from $(-3,0)$ to $(-3,-5)$ vertically, which is a shift 5 units down.

So, compared to $y = x^2$, the graph of $y = h(x)$ is shifted 3 units left and 5 units down.