These seven questions cover the linear law chapter in rising difficulty. Try each one on paper before opening the answer.
Questions
Question 1. Reduce y = 5x² + 2 to the form Y = mX + c. State Y, X, m and c.
Answer
The relation already has the shape y = a(x²) + b. So Y = y, X = x², m = 5, c = 2. When x = 3, X = 9 and Y = 5(9) + 2 = 47.
Question 2. The relation is y = px³ + qx. A graph of y/x against x² is a straight line through (1, 4) and (4, 19). Find p and q.
Answer
Divide by x: y/x = px² + q. So Y = y/x, X = x², m = p, c = q.
Gradient: (19 − 4) ÷ (4 − 1) = 5, so p = 5. Substitute (1, 4): 4 = 5 + c, so c = −1 and q = −1.
Check: y = 5x³ − x, and when x = 2, y = 38 and y/x = 19, which matches X = 4.
Question 3. The relation is y = a·b^x. A graph of lg y against x is a straight line through (0, 1.0) and (3, 1.6). Find a and b.
Answer
The linear form is lg y = (lg b)x + lg a. The intercept is 1.0, so a = 10^1.0 = 10.
Gradient: (1.6 − 1.0) ÷ 3 = 0.2, so lg b = 0.2 and b = 10^0.2 = 1.58 (3 s.f.).
Question 4. The relation is y = p/x + qx. Reduce it to linear form, then find p and q if the graph of xy against x² passes through (1, 7) and (9, 31).
Answer
Multiply by x: xy = p + qx². So Y = xy, X = x², m = q, c = p.
Gradient: (31 − 7) ÷ (9 − 1) = 3, so q = 3. Substitute (1, 7): 7 = 3 + c, so p = 4.
Check: y = 4/x + 3x. When x = 3, y = 4/3 + 9, so xy = 4 + 27 = 31. That matches X = 9.
Question 5. A graph of y against x² passes through (4, 10) and (16, 34). Find the relation, then find y when x = 5 and x when y = 74.
Answer
Gradient: (34 − 10) ÷ (16 − 4) = 2. Substitute (4, 10): 10 = 8 + c, so c = 2. The relation is y = 2x² + 2.
When x = 5: y = 2(25) + 2 = 52.
When y = 74: 2x² = 72, so x² = 36 and x = 6 (take the positive value if x is a length).
Question 6. A student says: “The graph of lg y against x has gradient 0.5, so in y = ab^x, b = 0.5.” Explain the error and give the correct b.
Answer
The gradient of lg y against x equals lg b, not b. So lg b = 0.5 and b = 10^0.5 = 3.16 (3 s.f.).
A check: b = 0.5 would make y fall as x grows, but a positive gradient in lg y means y grows.
Question 7. A graph of lg y against lg x is a straight line through (0, 0.40) and (0.6, 1.60). The relation is y = k·x^n. Find k and n, then y when x = 10.
Answer
Gradient: (1.60 − 0.40) ÷ 0.6 = 2, so n = 2.
Intercept 0.40 gives lg k = 0.40, so k = 10^0.40 = 2.51 (3 s.f.). The relation is y = 2.51x².
When x = 10: y = 2.51 × 100 = 251.
If you got these wrong
- Questions 1, 2 and 4 test reduction. Review turning a curve into a straight line.
- Questions 2, 3, 5 and 7 need the gradient and intercept. Review finding constants from gradient and intercept.
- Question 6 is about converting back. Review reading a linear-law model in context.
Record which step broke in the mistake log, and build a timed set with the timed original practice session builder. If you want a teacher to go through your working, see online one-to-one Additional Mathematics tuition.