Algebra 2 Transformations of Functions Worksheet Solutions (PDF)

Left to right, top to bottom: absolute value (vertex at the origin), square root (starting point at the origin), cube root (center at the origin), and quadratic (vertex at the origin).

Absolute value, square root, cube root, quadratic

\(x + 4\) shifts the graph left 4 and \(-2\) shifts it down 2.

Left 4, down 2; vertex (-4, -2)

Right 3: replace \(x\) with \(x – 3\). Up 5: add 5.

The radicand must be nonnegative: \(x – 3 \ge 0\).

\(g(x) = \sqrt{x – 3} + 5\), domain \(x \ge 3\)

\(-f(x)\) reflects over the x-axis: the graph goes down-right from the origin.

\(f(-x)\) reflects over the y-axis: the graph goes up-left, with domain \(x \le 0\).

g: reflection over the x-axis; h: reflection over the y-axis

Every output is multiplied by 3, a vertical stretch by a factor of 3.

\(g(1) = 3(1) = 3\); \(g(-2) = 3(2) = 6\).

Stretch by 3; g(1) = 3, g(-2) = 6

\(f(2x) = (2x)^{2} = 4x^{2} = 4f(x)\). For a quadratic, a horizontal compression by \(\frac{1}{2}\) looks the same as a vertical stretch by 4.

f(2x) = 4x² = 4 f(x)

Each point \((x, y)\) moves to \((x + 1, y + 2)\): \((-2, 4) \text{ becomes }(-1, 6)\) is written as the point \((-1, 6)\); \((0, 1)\) gives \((1, 3)\); \((1, -2)\) gives \((2, 0)\); \((3, 5)\) gives \((4, 7)\).

g: (-1, 6), (1, 3), (2, 0), (4, 7)

\(-f(x)\) changes the sign of \(y\): \((-2, -4)\), \((0, -1)\), \((1, 2)\), \((3, -5)\).

\(f(-x)\) changes the sign of \(x\): \((2, 4)\), \((0, 1)\), \((-1, -2)\), \((-3, 5)\).

h reflects over the x-axis; k reflects over the y-axis

The vertex moved from \((0, 0)\) to \((2, -3)\): right 2, down 3. The slopes are still \(\pm 1\).

\(g(x) = |x – 2| – 3\)

Left 1, vertical stretch by 2, reflection over the x-axis, up 3.

\(g(3) = -2\sqrt{4} + 3 = -4 + 3 = -1\).

Domain: \(x + 1 \ge 0\), so \(x \ge -1\). Since \(\sqrt{x + 1} \ge 0\), \(g(x) \le 3\).

g(3) = -1; domain x ≥ -1; range y ≤ 3

The center moves from \((0, 0)\) to \((1, -2)\).

\(g(9) = \sqrt[3]{8} – 2 = 2 – 2 = 0\).

Center (1, -2); g(9) = 0

\(f(-x) = (-x)^{4} – 3(-x)^{2} = x^{4} – 3x^{2} = f(x)\).

Even: symmetric about the y-axis

\(g(-x) = (-x)^{3} + (-x) = -x^{3} – x = -(x^{3} + x) = -g(x)\).

Odd: symmetric about the origin

\(h(1) = 2\) and \(h(-1) = 1 – 1 = 0\).

\(h(-1) \ne h(1)\) (not even) and \(h(-1) \ne -h(1)\) (not odd).

Neither

\(-3 < 1\): use \(x + 2\), so \(f(-3) = -1\). \(1 \ge 1\): use \(x^{2}\), so \(f(1) = 1\). \(f(4) = 16\).

-1, 1, 16

\(f(0) = -1\); \(f(2) = 2(2) – 1 = 3\); \(f(4) = 5 – 4 = 1\).

Check the other piece at \(x = 2\): \(5 – 2 = 3\), so the pieces meet.

f(0) = -1, f(2) = 3, f(4) = 1; no break

The graph of \(|t|\) is shifted right 68. \(D(75) = |7| = 7\); \(D(60) = |-8| = 8\).

Right 68; D(75) = 7 and D(60) = 8

Reflect over the x-axis, shift right 5, shift up 25. The vertex \((5, 25)\) is the maximum.

Ground: \(-(x – 5)^{2} + 25 = 0\), so \((x – 5)^{2} = 25\) and \(x = 0\) or \(10\).

Maximum 25 ft; the arch is 10 ft wide at the ground

For \(g\): \(\frac{x}{2} = 2\) gives \(x = 4\), so \((4, 3)\). For \(h\): \(2x = 2\) gives \(x = 1\), so \((1, 3)\).

g: (4, 3); h: (1, 3)

\(f(x + 3)\) gives the same output 3 units earlier. If \(f(0) = 5\), then \(g(-3) = f(0) = 5\), so the point moves from \(x = 0\) to \(x = -3\).

The graph shifts left 3

Solve \(2x – 4 = 6\): \(x = 5\). The output is unchanged.

(5, 5)