Transformations of Graphs: Additional Practice (Set 1)

The point reappears where the inside equals the old input $9$.

$3x + 12 = 9 \ \Rightarrow\ 3x = -3 \ \Rightarrow\ x = -1$

Solve the inside for $x$; the height $4$ is unchanged.

Compute the compression and compare with a vertical stretch.

$f(3x) = (3x)^2 = 9x^2 = 9f(x)$

So it equals a vertical stretch by $9$. This holds for any homogeneous power $x^d$, where $f(bx) = b^d f(x)$, but it fails once a constant is added.

Expand the second and subtract the first.

$3(f(x) + 2) - (3f(x) + 2) = (3f(x) + 6) - (3f(x) + 2) = 4$

The stretch multiplied the $2$ into a $6$, leaving a gap of $4$.

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

$f(x) \ \to\ 3f(x) \ \to\ 3f(x) - 5$

Stretch first, so the $5$ is subtracted afterward and is not multiplied.

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

$f(x) \ \to\ f(x) - 5 \ \to\ 3(f(x) - 5) = 3f(x) - 15$

Shift first, so the later stretch triples the $-5$ into $-15$.

Even means the input reflection changes nothing.

$g(-x) = g(x)$

So $y = g(-x)$ is the same graph; the reflection leaves it unchanged.

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

$x^2 \to 4x^2 = 4f, \quad x^4 \to 16x^4 = 16f, \quad 2x^2 \to 8x^2 = 4f, \quad x^2 - 3 \to 4x^2 - 3$

Only $4x^2 - 3$ is not a multiple of $f(x) = x^2 - 3$, so that one fails.

The point reappears where the inside equals the old input $2$.

$\tfrac{1}{3}x - 1 = 2 \ \Rightarrow\ \tfrac{1}{3}x = 3 \ \Rightarrow\ x = 9$

Solve the inside for $x$; the height $6$ is unchanged.

The inside $3x$ reaches the old inputs at a third of the $x$, so the domain is divided by $3$.

$-6 \le 3x \le 9 \ \Rightarrow\ -2 \le x \le 3$

Each endpoint is divided by $3$.

Multiplying by $-2$ scales the outputs and, being negative, swaps their order.

$-2 \times (-1) = 2, \qquad -2 \times 4 = -8$

The endpoints become $2$ and $-8$, so the range runs $-8 \le y \le 2$.

The inside $-x$ reflects across the $y$-axis, and the outside $+3$ shifts up.

$f(x) \ \to\ f(-x) \ \to\ f(-x) + 3$

Negating the input is a $y$-axis reflection, not an $x$-axis one, and $+3$ raises the graph.

Both a horizontal and a vertical move reach the same graph.

$f(3x) = 9x^2 = 9f(x)$

So the compression by $3$ and the stretch by $9$ produce the identical graph, and the graph does not record which was used.