Shifting Graphs: Practice

The graph $y = f(x + 2) - 3$ is a shift left $2$ and down $3$, so subtract $2$ from the input and $3$ from the output.

$(p, q) \longrightarrow (p - 2,\; q - 3)$

Vertex $(-2, 5)$ gives $h = -2$ (so the inside is $x + 2$) and $k = 5$ in $y = (x - h)^2 + k$.

$y = (x + 2)^2 + 5$

The inside $x + 2$ is a left shift by $2$, which moves both x-intercepts left by $2$.

$(1 - 2,\; 0) = (-1, 0), \qquad (4 - 2,\; 0) = (2, 0)$

At the shared input $3$, the second graph is the first raised by $c$, so its height is $f(3) + c$.

$2 + c = 9 \;\Rightarrow\; c = 7$

Set the output to $0$ and solve for $x$.

$(x - 1)^2 = 9 \;\Rightarrow\; x - 1 = \pm 3 \;\Rightarrow\; x = 4 \text{ or } x = -2$

The crossings are $(4, 0)$ and $(-2, 0)$, symmetric about the vertex input $x = 1$.

Left $1$ shifts the domain left $1$, and down $4$ shifts the range down $4$.

$-3 - 1 \le x \le 2 - 1, \qquad 0 - 4 \le y \le 8 - 4$

That gives domain $-4 \le x \le 1$ and range $-4 \le y \le 4$.

With $g(x) = f(x - 2) + 1$, at $x = 5$ the inside is $5 - 2 = 3$.

$g(5) = f(3) + 1 = 10 \;\Rightarrow\; f(3) = 9$

The height is unchanged and the input grows from $2$ to $9$, so the horizontal shift is their difference.

$h = 9 - 2 = 7$

The vertex moves from $(0, 0)$ to $(6, -4)$, so the input rose by $6$ and the height fell by $4$.

$(0, 0) \longrightarrow (6, -4) \;\Rightarrow\; \text{right } 6,\ \text{down } 4$

The equation is $y = (x + 1)^2 - 6$. Put $x = 0$ to find the y-intercept.

$(0 + 1)^2 - 6 = 1 - 6 = -5 \;\Rightarrow\; (0, -5)$

The input is unchanged and the output falls from $10$ to $3$, so the downward shift is their difference.

$k = 10 - 3 = 7$

Points moving to $(x - 3, y + 2)$ means the graph shifts left $3$ and up $2$. Left $3$ makes the inside $x + 3$; up $2$ adds $2$ outside.

$y = f(x + 3) + 2$