Transformations of Graphs: Practice

Adding $5$ outside the function is a vertical shift, so keep the input and add $5$ to the height.

$(4,\ 7 + 5) = (4, 12)$

The input $4$ does not change, and the height rises by $5$.

The inside change $x - 3$ shifts the graph right by $3$, so add $3$ to the input and keep the height.

$(6 + 3,\ 2) = (9, 2)$

The minus sign inside moves the graph right, not left.

Multiplying by $3$ outside is a vertical stretch, so keep the input and multiply the height by $3$.

$(2,\ 3 \times 5) = (2, 15)$

The input is unchanged; only the height scales.

An inside change shifts the graph horizontally, and $x - 6$ moves it in the direction opposite the sign.

$y = f(x - 6) \ \Rightarrow\ \text{right } 6$

Subtracting inside moves the graph right, because the inside reaches an old value only when $x$ grows.

Reflecting across the $x$-axis negates every height, which is a minus sign on the output.

$y = -f(x)$

Negating the input, $f(-x)$, would reflect across the $y$-axis instead.

Reflecting across the $y$-axis negates every input, which is a minus sign inside the function.

$y = f(-x)$

Negating the output, $-f(x)$, would reflect across the $x$-axis instead.

The rule $f(-x)$ reflects across the $y$-axis, negating the input while the height stays.

$(-3,\ 8)$

The height $8$ is unchanged; only the input flips sign.

The rule $-f(x)$ reflects across the $x$-axis, negating the height while the input stays.

$(-5,\ -6)$

The input $-5$ is unchanged; only the height flips sign.

The point reappears where the inside $\tfrac{x}{2}$ equals the old input $8$.

$\tfrac{x}{2} = 8 \ \Rightarrow\ x = 16, \quad \text{height } 3$

The inside multiplier $\tfrac{1}{2}$ is below $1$, so the graph stretches horizontally and every input doubles.

Adding $k$ raises every height by $k$, so a downward move needs a negative amount.

$k < 0 \ \Rightarrow\ \text{shift down}$

A positive $k$ shifts up and $k = 0$ leaves the graph in place.

Adding $3$ outside raises the height by $3$ while the input stays.

$(2,\ -3 + 3) = (2, 0)$

The new height is $0$, not $3$: you add to the old height, not replace it.

Multiplying outside by $\tfrac{1}{2}$ is a vertical compression, halving each height.

$\left(10,\ \tfrac{1}{2} \times 4\right) = (10, 2)$

The input is unchanged; only the height scales.