Shifting Graphs: Additional Practice (Set 2)

An x-intercept is a point on the graph, so it moves with the graph. A right shift by $5$ adds $5$ to its input.

$(2 + 5,\; 0) = (7, 0)$

The y-intercept is the value at $x = 0$.

$(0 - 4)^2 + 2 = 16 + 2 = 18 \;\Rightarrow\; (0, 18)$

The vertex is at $(4, 2)$, but the y-intercept is far higher because $x = 0$ is four units from the vertex.

The known point moves right $4$ and up $1$ with the graph.

$(0 + 4,\; 5 + 1) = (4, 6)$

After a horizontal shift the y-intercept is not guaranteed to stay put, so track the given point directly.

To send $(0, 0)$ to $(3, 3)$, move right $3$ and up $3$. Right $3$ makes the inside $x - 3$; up $3$ adds $3$ outside.

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

Left $3$ and up $1$ means $g(x) = f(x + 3) + 1$. At $x = 1$ the inside is $1 + 3 = 4$.

$g(1) = f(4) + 1 = 7 + 1 = 8$

Both are shifted up $7$, so compare only the horizontal moves: the first is right $2$, the second is left $2$.

$2 - (-2) = 4 \;\Rightarrow\; \text{the first is } 4 \text{ units to the right}$

The rule shifts right $h$ (input increases by $h$) and up $k$ (output increases by $k$).

$(a, b) \longrightarrow (a + h,\; b + k)$

Each end of the range falls by the same amount: $3 \to -1$ and $12 \to 8$, a drop of $4$.

$y = f(x) - 4 \;\Rightarrow\; \text{down } 4$

The inside $x - 6$ is a right shift by $6$, so the peak moves right $6$ with its height unchanged.

$(2 + 6,\; 5) = (8, 5)$

Both share $k = -3$, so compare the vertices: $(-2, -3)$ and $(5, -3)$.

$5 - (-2) = 7 \;\Rightarrow\; \text{right } 7$

From $x + 4$ and $+1$, the vertex is $(-4, 1)$. The parabola opens upward with lowest height $1 > 0$, so it never reaches the axis.

$\text{lowest } y = 1 > 0 \;\Rightarrow\; \text{no x-intercepts}$

Undo the shift: the image moved right $h$ and up $k$, so the original point was left $h$ and down $k$.

$(a - h,\; b - k)$