Transformations of Graphs: Practice

The point reappears where the inside equals the old input $8$.

$2x - 6 = 8 \ \Rightarrow\ 2x = 14 \ \Rightarrow\ x = 7$

Solve the inside for $x$; the height $5$ is unchanged.

Compute both and try to match them.

$f(2x) = 4x^2 + 1, \qquad c\,f(x) = c x^2 + c$

Matching the $x^2$ term needs $c = 4$, but matching the constant needs $c = 1$, so no single $c$ works. The added constant breaks the coincidence.

Apply the moves in order to the output, shift first and then stretch.

$f(x) \ \to\ f(x) + 3 \ \to\ 2(f(x) + 3) = 2f(x) + 6$

The shift happens first, so the later stretch doubles the $3$ into a $6$.

Even means the input reflection changes nothing.

$f(-x) = f(x)$

So the graph of $y = f(-x)$ is identical to the graph of $y = f(x)$; evenness is invariance under reflection across the $y$-axis.

Odd means reflecting the input matches reflecting the output.

$f(-x) = -f(x)$

So the input reflection $y = f(-x)$ is the same graph as the output reflection $y = -f(x)$, a reflection across the $x$-axis.

Test each: a pure power scales, a power plus a constant does not.

$x^2 \to 4x^2 = 4f, \quad x^3 \to 8x^3 = 8f, \quad 5x^2 \to 20x^2 = 4f, \quad x^2 + 1 \to 4x^2 + 1$

Only $4x^2 + 1$ is not a multiple of $f(x) = x^2 + 1$, so that one fails to be a vertical stretch.

The point reappears where the inside equals the old input $3$.

$\tfrac{1}{2}x + 1 = 3 \ \Rightarrow\ \tfrac{1}{2}x = 2 \ \Rightarrow\ x = 4$

Solve the inside for $x$; the height $10$ is unchanged.

Add $5$, then negate the height; the input is unchanged.

$(2, 3) \ \to\ (2, 8) \ \to\ (2,\ -8)$

Here the shift happens first, so reflection negates the raised height. The other order gives $(2, 2)$, so order matters.

Every input is divided by $b$, and $0 < b < 1$ makes $\tfrac{1}{b} > 1$.

$0 < b < 1 \ \Rightarrow\ \tfrac{1}{b} > 1$

Inputs are pushed away from the $y$-axis, a horizontal stretch.

Check which single moves give $4x^2$.

$f(2x) = 4x^2 \quad\text{and}\quad 4f(x) = 4x^2$

Compression by $2$ and stretch by $4$ both give $4x^2$. A vertical compression by $4$ gives $\tfrac{1}{4}x^2$, a horizontal stretch by $2$ gives $\tfrac{1}{4}x^2$, and a compression by $4$ gives $16x^2$, so only that one pair works.

Factor the inside into the form $b(x - h)$.

$2x - 4 = 2(x - 2)$

So $h = 2$: the shift is right $2$. The un-factored reading 'right $4$' is the trap.

Shift the input right $2$; scale the height by $3$ then subtract $4$.

$(6 + 2,\ 3 \times 1 - 4) = (8, -1)$

Multiply before subtracting: $3 \times 1 = 3$, then $3 - 4 = -1$.