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Quiz: Linear Functions

Test your understanding of linear functions, gradients, equation forms, and real-world linear models with these review questions.

1. What is the gradient of the line passing through \((2, 5)\) and \((6, 13)\)?

  1. \(\frac{1}{2}\)
  2. \(2\)
  3. \(4\)
  4. \(\frac{9}{4}\)
Show Answer

The correct answer is B. Using the gradient formula: \(m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\). The gradient tells us that for every \(1\) unit increase in \(x\), \(y\) increases by \(2\). A common mistake is to invert the formula or mix up the subtraction order — always keep the subtraction in the numerator and denominator consistent.

Concept Tested: Gradient Calculation

2. A line is perpendicular to \(y = \frac{2}{3}x + 5\). What is its gradient?

  1. \(\frac{2}{3}\)
  2. \(-\frac{2}{3}\)
  3. \(\frac{3}{2}\)
  4. \(-\frac{3}{2}\)
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The correct answer is D. Perpendicular lines have gradients whose product is \(-1\). The negative reciprocal of \(\frac{2}{3}\) is \(-\frac{3}{2}\). Check: \(\frac{2}{3} \times -\frac{3}{2} = -1\). A common mistake is to take just the reciprocal (\(\frac{3}{2}\)) without the negative sign, or to take just the negative (\(-\frac{2}{3}\)) without inverting.

Concept Tested: Perpendicular Lines

3. Which equation represents a line in gradient-intercept form?

  1. \(y = -3x + 4\)
  2. \(2x + 3y - 6 = 0\)
  3. \(y - 1 = 2(x - 3)\)
  4. \(x = 5\)
Show Answer

The correct answer is A. Gradient-intercept form is \(y = mx + c\), where \(m\) is the gradient and \(c\) is the \(y\)-intercept. Option A matches this pattern with \(m = -3\) and \(c = 4\). Option B is general form, option C is point-gradient form, and option D is a vertical line, which is not a function and cannot be written in gradient-intercept form.

Concept Tested: Gradient-Intercept Form

4. A line has gradient \(-4\) and passes through the point \((2, 3)\). What is its equation in gradient-intercept form?

  1. \(y = -4x + 3\)
  2. \(y = -4x + 11\)
  3. \(y = -4x - 5\)
  4. \(y = -4x + 14\)
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The correct answer is B. Start with point-gradient form: \(y - 3 = -4(x - 2)\). Expand: \(y - 3 = -4x + 8\). Add \(3\) to both sides: \(y = -4x + 11\). Verify by substituting \((2, 3)\): \(-4(2) + 11 = -8 + 11 = 3\). ✓ A common mistake is dropping the sign when distributing \(-4\) across \((x - 2)\).

Concept Tested: Point-Gradient Form

5. Which statement about a horizontal line is true?

  1. It has an undefined gradient
  2. It has a positive gradient
  3. It has a zero gradient
  4. It is not a function
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The correct answer is C. A horizontal line neither rises nor falls, so its rise is zero for any run, giving \(m = 0\). Undefined gradients belong to vertical lines (which are not functions). Horizontal lines like \(y = 4\) are perfectly valid functions — they just happen to be constant functions. Every input produces the same output.

Concept Tested: Zero Gradient

6. Find the intersection point of \(y = 2x + 1\) and \(y = -x + 7\).

  1. \((2, 5)\)
  2. \((3, 7)\)
  3. \((1, 3)\)
  4. \((4, 3)\)
Show Answer

The correct answer is A. Set the expressions equal: \(2x + 1 = -x + 7\), so \(3x = 6\) and \(x = 2\). Substitute back into either equation: \(y = 2(2) + 1 = 5\). The intersection is \((2, 5)\). Always verify by checking that both equations give the same \(y\) at that \(x\): \(-2 + 7 = 5\). ✓

Concept Tested: Intersection of Lines

7. A taxi company charges a \(\$4\) flag fall plus \(\$2.50\) per kilometre. Which function gives the total cost \(C\) for a journey of \(k\) kilometres?

  1. \(C(k) = 4k + 2.50\)
  2. \(C(k) = 2.50k + 4\)
  3. \(C(k) = 6.50k\)
  4. \(C(k) = 4 - 2.50k\)
Show Answer

The correct answer is B. The rate of change is \(\$2.50\) per kilometre (this is the gradient, multiplying \(k\)), and the starting value is \(\$4\) when no distance has been traveled (this is the \(y\)-intercept). So \(C(k) = 2.50k + 4\). A common error is swapping the gradient and \(y\)-intercept (option A), which would mean charging \(\$4\)/km with a \(\$2.50\) fixed fee.

Concept Tested: Real-World Linear Models

8. Two lines are parallel. Line 1 has equation \(y = 5x - 2\). Which could be Line 2?

  1. \(y = -5x + 3\)
  2. \(y = \frac{1}{5}x + 1\)
  3. \(y = 5x + 7\)
  4. \(y = -\frac{1}{5}x - 2\)
Show Answer

The correct answer is C. Parallel lines have identical gradients. Line 1 has gradient \(5\), so any parallel line must also have gradient \(5\). Option C has gradient \(5\) and a different \(y\)-intercept, meaning it runs alongside Line 1 without ever crossing it. Options A and D show reflected/reciprocal variants, and option B would be perpendicular if multiplied correctly — but neither matches gradient \(5\).

Concept Tested: Parallel Lines

9. Which relationship represents direct proportion?

  1. \(y = 3x + 2\)
  2. \(y = x^2\)
  3. \(y = 4x\)
  4. \(y = \frac{5}{x}\)
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The correct answer is C. Direct proportion has the form \(y = kx\), where \(k\) is a non-zero constant and the graph passes through the origin \((0, 0)\). Option C fits exactly with \(k = 4\). Option A is linear but not direct proportion (the \(y\)-intercept is \(2\), not \(0\)). Option B is quadratic, and option D is inverse proportion.

Concept Tested: Direct Proportion

10. Solve for \(x\): \(7x - 4 = 3x + 12\).

  1. \(x = 2\)
  2. \(x = 3\)
  3. \(x = 8\)
  4. \(x = 4\)
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The correct answer is D. Collect \(x\) terms on one side and constants on the other: \(7x - 3x = 12 + 4\), giving \(4x = 16\), so \(x = 4\). Check: \(7(4) - 4 = 24\) and \(3(4) + 12 = 24\). ✓ The key is to keep the equation balanced — whatever you do to one side, do to the other.

Concept Tested: Linear Equation Solving