Determining Constants from a Linear Graph

Question 1

To reduce y = axⁿ to linear form, you plot:

Accepted Answer

lg y against lg x — lg y = n lg x + lg a is linear in lg x, so plot lg y against lg x.

Question 2

For y = abˣ, the graph of lg y against x is a straight line with gradient:

Accepted Answer

lg b — lg y = (lg b)x + lg a, so the gradient is lg b.

Question 3

For y = abˣ, the vertical intercept of the graph of lg y against x gives:

Accepted Answer

lg a — When x = 0, lg y = lg a, so the intercept equals lg a.

Question 4

For y = axⁿ, a graph of lg y against lg x has gradient 2 and intercept 3. Then:

Accepted Answer

a = 1000, n = 2 — n = gradient = 2 and lg a = intercept = 3, so a = 10³ = 1000.

Question 5

A graph of y against x² is a straight line with gradient 4 and intercept 5. The relationship is:

Accepted Answer

y = 4x² + 5 — With Y = y and X = x², y = 4x² + 5 (gradient 4, intercept 5).

Question 6

To obtain a straight line from y = a/x + b, plot y against:

Accepted Answer

1/x — y = a(1/x) + b is linear in 1/x with gradient a and intercept b.

Question 7

For y = abˣ, the graph of lg y against x has intercept 2. The value of a is:

Accepted Answer

100 — Intercept = lg a = 2, so a = 10² = 100.

Question 8

The relationship y = px + qx² can be reduced to a straight line by plotting:

Accepted Answer

y/x against x — Dividing by x gives y/x = p + qx, linear in x with gradient q and intercept p.

Question 9

For y = kxⁿ, a graph of lg y against lg x passes through (0, 1) with gradient 3. Then:

Accepted Answer

k = 10, n = 3 — n = gradient = 3, and lg k = intercept = 1, so k = 10¹ = 10.

Question 10

The relationship y = a√x becomes a straight line if you plot y against:

Accepted Answer

√x, with gradient a — y = a√x is linear in √x; plotting y against √x gives a line through the origin of gradient a.