Unit 10 · Checkpoint quiz

Unit 10 checkpoint: Functions

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next.

If \(f(x) = -x^2 + 3x - 4\), find \(f(-2)\).

Replace every \(x\) with \((-2)\), in parentheses: \(f(-2) = -(-2)^2 + 3(-2) - 4\).

Exponent first: \((-2)^2 = 4\), so \(-(-2)^2 = -4\).

\(-4 - 6 - 4 = -14\). Answer: \(-14\).

If \(f(x) = 2x^2 - x\), find \(f(a + 1)\) and simplify.

Replace \(x\) with \((a + 1)\): \(f(a + 1) = 2(a + 1)^2 - (a + 1)\).

\((a + 1)^2 = a^2 + 2a + 1\), so \(2(a + 1)^2 = 2a^2 + 4a + 2\).

\(2a^2 + 4a + 2 - a - 1 = 2a^2 + 3a + 1\). Answer: \(2a^2 + 3a + 1\).

Which relation is not a function?

  1. \(\{(1, 2), (2, 3), (3, 4)\}\)
  2. \(\{(1, 5), (2, 5), (3, 5)\}\)
  3. \(\{(2, 1), (2, 3), (4, 5)\}\)
  4. \(\{(0, 0), (1, 1), (-1, 1)\}\)

A function gives each input (\(x\)) exactly one output (\(y\)).

In c, the input 2 has two outputs, 1 and 3. So c is not a function.

In b and d, different inputs share an output. That is allowed. Answer: \(\{(2, 1), (2, 3), (4, 5)\}\).

Find the domain of \(f(x) = \sqrt{6 - 2x}\). Write it in interval notation.

An even root needs the inside to be zero or positive: \(6 - 2x \ge 0\).

Subtract 6: \(-2x \ge -6\). Divide by \(-2\) and flip the sign: \(x \le 3\).

Answer: \((-\infty, 3]\).

Find the domain of \(g(x) = \dfrac{x + 1}{x^2 - 4x}\). Write it in interval notation.

The denominator cannot be 0: \(x^2 - 4x = x(x - 4) = 0\) when \(x = 0\) or \(x = 4\).

Every other real number is allowed. Leave out 0 and 4 with round brackets.

Answer: \((-\infty, 0) \cup (0, 4) \cup (4, \infty)\).

The graph below shows the whole function (it stops at the two dots). What is its range? Write it in interval notation.

The graph starts at \((-3, 1)\), goes down to \((0, -2)\), and ends at \((4, 2)\). Both end dots are filled in.

Range means all the \(y\)-values the graph reaches.

The lowest point is \((0, -2)\), so the smallest \(y\) is \(-2\). The highest point is the end \((4, 2)\), so the largest \(y\) is \(2\).

The end dots are filled, so both are included. Answer: \([-2, 2]\).

How is the graph of \(g(x) = -(x + 3)^2 + 1\) made from the graph of \(y = x^2\)?

  1. Reflect over the x-axis, then move right 3 and up 1
  2. Reflect over the x-axis, then move left 3 and up 1
  3. Reflect over the y-axis, then move left 3 and up 1
  4. Reflect over the x-axis, then move left 3 and down 1

The minus sign in front flips the parabola upside down: a reflection over the x-axis.

\(x + 3\) inside means a shift 3 to the left (the opposite of the sign you see).

\(+1\) outside means up 1. Answer: reflect over the x-axis, then move left 3 and up 1.

The graph of \(y = \sqrt{x}\) is moved 3 units left and 4 units down. Write the equation of the new graph.

Left 3 changes \(x\) to \(x + 3\) inside the root: \(y = \sqrt{x + 3}\).

Down 4 subtracts 4 outside: \(y = \sqrt{x + 3} - 4\).

Answer: \(y = \sqrt{x + 3} - 4\).

Let \(f(x) = 2x - 5\) and \(g(x) = x^2 + 1\). Find \((f \circ g)(3)\).

\((f \circ g)(3) = f(g(3))\): work from the inside out.

\(g(3) = 3^2 + 1 = 10\).

\(f(10) = 2(10) - 5 = 15\). Answer: \(15\).

Find the inverse of \(f(x) = \dfrac{3x - 2}{5}\).

Write \(y = \dfrac{3x - 2}{5}\), then swap \(x\) and \(y\): \(x = \dfrac{3y - 2}{5}\).

Solve for \(y\): multiply by 5 to get \(5x = 3y - 2\); add 2 to get \(5x + 2 = 3y\).

Divide by 3. Answer: \(f^{-1}(x) = \dfrac{5x + 2}{3}\).

Check: \(f(f^{-1}(x)) = \dfrac{3 \cdot \frac{5x + 2}{3} - 2}{5} = \dfrac{5x + 2 - 2}{5} = x\).

Solve \(2x^2 = 7x + 15\).

Standard form (everything on one side, 0 on the other): \(2x^2 - 7x - 15 = 0\).

Factor: \((2x + 3)(x - 5) = 0\). (Check: \(2x^2 - 10x + 3x - 15\).)

Zero product property: \(2x + 3 = 0\) gives \(x = -\dfrac{3}{2}\); \(x - 5 = 0\) gives \(x = 5\).

Answer: \(x = 5\) or \(x = -\dfrac{3}{2}\).

Solve \(\dfrac{x}{3} - \dfrac{x - 2}{4} = 2\).

The LCD is 12. Multiply every term by 12: \(4x - 3(x - 2) = 24\).

Distribute the \(-3\): \(4x - 3x + 6 = 24\), so \(x + 6 = 24\).

\(x = 18\). Check: \(\dfrac{18}{3} - \dfrac{16}{4} = 6 - 4 = 2\). Answer: \(x = 18\).

Want to look something up first? Functions cheat sheet · Glossary

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