Exercise 10.1
Exercise 10.1 Solutions
Plotting and sketching graphs of linear, quadratic, cubic, exponential, reciprocal, and radical functions.
Question 1
Linear Graphs
Sketch the graph of the following linear functions:
(i) y = 3x - 5 (ii) y = -2x + 8 (iii) y = 0.5x - 1
(i) y = 3x - 5 (ii) y = -2x + 8 (iii) y = 0.5x - 1
Solution Step-by-Step
(i) y = 3x - 5
Substitute values for x to find corresponding y values:
| x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | -8 | -5 | -2 | 1 |
| (x, y) | (-1, -8) | (0, -5) | (1, -2) | (2, 1) |
Plot these coordinates on the grid. Join the points with a straight line. The scale on the x-axis is 1 division = 1 unit, and on the y-axis is 1 division = 2 units. The line has y-intercept at y = -5 and slope m = 3.
(ii) y = -2x + 8
Substitute values for x to find corresponding y values:
| x | 0 | 2 | 4 |
|---|---|---|---|
| y | 8 | 4 | 0 |
| (x, y) | (0, 8) | (2, 4) | (4, 0) |
Plot coordinates and join them. The line has y-intercept at y = 8, x-intercept at x = 4, and slope m = -2 (slanted downwards).
(iii) y = 0.5x - 1
Substitute values:
| x | -2 | 0 | 2 | 4 |
|---|---|---|---|---|
| y | -2 | -1 | 0 | 1 |
| (x, y) | (-2, -2) | (0, -1) | (2, 0) | (4, 1) |
Join points. y-intercept is y = -1 and x-intercept is x = 2. Slope is 0.5 (gentle upward slant).
Question 2
Quad & Cubic Graphs
Sketch the graph of the following quadratic and cubic functions:
(i) y = x³ + 2x² - 5x - 6 for -3.5 ≤ x ≤ 2.5
(ii) y = x² + x - 2
(iii) y = x³ + 3x² + 2x for -2.5 ≤ x ≤ 0.5
(iv) y = 5x² - 2x - 3
(i) y = x³ + 2x² - 5x - 6 for -3.5 ≤ x ≤ 2.5
(ii) y = x² + x - 2
(iii) y = x³ + 3x² + 2x for -2.5 ≤ x ≤ 0.5
(iv) y = 5x² - 2x - 3
Solution Step-by-Step
(i) y = x³ + 2x² - 5x - 6 (Cubic)
Calculate coordinates for the given interval:
| x | -3.5 | -3 | -2 | -1 | 0 | 1 | 2 | 2.5 |
|---|---|---|---|---|---|---|---|---|
| y | -6.88 | 0 | 4 | 0 | -6 | -8 | 0 | 9.63 |
Join the plotted points with a smooth curve. It has turning points, cuts the x-axis at three roots: x = -3, x = -1, and x = 2, and cuts the y-axis at y = -6.
(ii) y = x² + x - 2 (Quadratic)
Calculate coordinates:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 4 | 0 | -2 | -2 | 0 | 4 | 10 |
This is a parabola opening upward (since coefficient of x² is positive). Vertex is at x = -0.5, y = -2.25. The roots are x = -2 and x = 1.
(iii) y = x³ + 3x² + 2x (Cubic)
Calculate coordinates:
| x | -2.5 | -2 | -1.5 | -1 | -0.5 | 0 | 0.5 |
|---|---|---|---|---|---|---|---|
| y | -1.88 | 0 | 0.38 | 0 | -0.38 | 0 | 1.88 |
Plot points and draw a smooth "S"-like cubic curve. The roots are x = -2, x = -1, and x = 0.
(iv) y = 5x² - 2x - 3 (Quadratic)
Calculate coordinates:
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| y | 21 | 4 | -3 | 0 | 13 |
Parabola opening upward. The y-intercept is y = -3 and it cuts the x-axis at x = 1 and x = -0.6.
Question 3
Other Functions
Sketch the graph of the following functions:
(i) y = 4x (ii) y = 5-x (iii) y = 1/(x-3) for x ≠ 3
(iv) y = 4/x + 3 for x ≠ 0 (v) y = √x (vi) y = 3x1/3
(vii) y = 2x²
(i) y = 4x (ii) y = 5-x (iii) y = 1/(x-3) for x ≠ 3
(iv) y = 4/x + 3 for x ≠ 0 (v) y = √x (vi) y = 3x1/3
(vii) y = 2x²
Solution Step-by-Step
(i) y = 4x (Exponential Growth)
Calculate coordinates:
| x | -2 | -1 | 0 | 0.5 | 1 | 2 |
|---|---|---|---|---|---|---|
| y | 0.06 | 0.25 | 1 | 2 | 4 | 16 |
As x increases, y increases rapidly. As x decreases, y approaches 0 (the x-axis is a horizontal asymptote).
(ii) y = 5-x (Exponential Decay)
Calculate coordinates:
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| y | 25 | 5 | 1 | 0.2 | 0.04 |
As x increases, the function decays towards 0. Y-intercept is at (0, 1).
(iii) y = 1/(x-3) for x ≠ 3 (Reciprocal)
Calculate coordinates around the asymptote x = 3:
| x | -3 | -1 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|
| y | -0.17 | -0.25 | -0.5 | -1 | undef | 1 | 0.5 | 0.33 |
The line x = 3 is a vertical asymptote. As x approaches 3 from the left, y goes to -∞. As it approaches from the right, y goes to +∞. Horizontal asymptote is y = 0.
(iv) y = 4/x + 3 for x ≠ 0 (Reciprocal)
Calculate coordinates around the asymptote x = 0:
| x | -4 | -2 | -1 | 0 | 1 | 2 | 4 |
|---|---|---|---|---|---|---|---|
| y | 2 | 1 | -1 | undef | 7 | 5 | 4 |
Vertical asymptote is x = 0 (y-axis). The horizontal asymptote is shifted up to y = 3.
(v) y = √x (Square Root)
Calculate coordinates:
| x | 0 | 1 | 4 | 9 | 16 | 25 | 36 |
|---|---|---|---|---|---|---|---|
| y | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
The function is not defined for negative values of x. The graph starts at origin (0, 0) and curves slowly upwards.
(vi) y = 3x1/3 (Cube Root)
Calculate coordinates:
| x | -8 | -1 | 0 | 1 | 8 |
|---|---|---|---|---|---|
| y | -6 | -3 | 0 | 3 | 6 |
Unlike square roots, cube roots are defined for negative values. The curve passes through the origin with a vertical tangent at (0, 0).
(vii) y = 2x² (Quadratic)
Calculate coordinates:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 18 | 8 | 2 | 0 | 2 | 8 | 18 |
A vertical stretch of the standard parabola by a factor of 2. Vertex is at the origin (0, 0).
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