KCSE 2025 Maths P2 Q20 — Cubic Graph, Table of Values & Solving by a Line

KCSE 2025 Form 4 Graphs

Published

The Question

“A table shows values of x and some values of y for the curve y = 5 + 10x - 2x squared - 4x cubed for x from -2 to 2, at x = -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2. (a) Complete the table by filling in the missing y values. (b) Using a scale of 2 cm for 1 unit on the x-axis and 2 cm for 5 units on the y-axis, draw the graph of y = 5 + 10x - 2x squared - 4x cubed. (c)(i) Use the graph to solve the equation 5 + 10x - 2x squared - 4x cubed = 0. (c)(ii) By drawing a suitable straight line on the graph, solve the equation 4x cubed + 2x squared - 12x - 3 = 0.”

1

Complete the table (part a)

For each missing x, substitute it into the curve and work out the powers carefully, watching the signs of the cube term. Doing this for the four missing entries at x = -2, x = -0.5, x = 0.5 and x = 2 fills the gaps. The x squared term is always positive and the cube term flips sign with x.

x=−2:  5+10(−2)−2(−2)2−4(−2)3=5−20−8+32=9x = -2:\; 5 + 10(-2) - 2(-2)^{2} - 4(-2)^{3} = 5 - 20 - 8 + 32 = 9
x=−0.5:  5−5−2(0.25)+4(0.125)=0x = -0.5:\; 5 - 5 - 2(0.25) + 4(0.125) = 0
x=0.5:  5+5−2(0.25)−4(0.125)=9x = 0.5:\; 5 + 5 - 2(0.25) - 4(0.125) = 9
x=2:  5+20−8−32=−5x = 2:\; 5 + 20 - 8 - 32 = -5

Completed y-row for x = -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2 is: 9, -1, -3, 0, 5, 9, 9, 2, -5.

2

Plot the curve (part b)

Draw the axes using the given scale, 2 cm for 1 unit across and 2 cm for 5 units up, then plot each point from the completed table. Join them with a single smooth curve, not straight segments. The cubic rises from the left, turns over near the middle and then falls steeply on the right.

3

Solve the curve = 0 by reading roots (part c i)

The equation to solve is exactly the curve set equal to zero, and y is zero everywhere on the x-axis. So the solutions are simply the x-values where the drawn curve crosses the x-axis. Read all three crossings straight down to the axis.

5+10x−2x2−4x3=05 + 10x - 2x^{2} - 4x^{3} = 0
x≈−1.58,x≈−0.5,x≈1.58x \approx -1.58,\quad x \approx -0.5,\quad x \approx 1.58
4

Turn the second cubic into a line (part c ii)

You must solve a different cubic using the same curve, so make the new equation share the curve's expression. From the curve, rearrange to get 4x cubed plus 2x squared on its own in terms of x and y. Substitute that block into the new equation wherever 4x cubed plus 2x squared appears, and the cubic terms collapse, leaving a straight line.

y=5+10x−2x2−4x3⇒4x3+2x2=5+10x−yy = 5 + 10x - 2x^{2} - 4x^{3} \Rightarrow 4x^{3} + 2x^{2} = 5 + 10x - y
(5+10x−y)−12x−3=0(5 + 10x - y) - 12x - 3 = 0
2−2x−y=0⇒y=2−2x2 - 2x - y = 0 \Rightarrow y = 2 - 2x
5

Draw the line and read the solutions

Draw the straight line y = 2 - 2x across the same axes; it has a y-intercept of 2 and slope -2. Wherever this line meets the curve, the x-values are the solutions of the new cubic, because at those points the curve and line share the same y. Read the three intersection points down to the x-axis.

x≈−1.85,x≈−0.25,x≈1.63x \approx -1.85,\quad x \approx -0.25,\quad x \approx 1.63

Final Result

The completed y-values are 9, -1, -3, 0, 5, 9, 9, 2, -5. The curve = 0 has roots x is about -1.58, -0.5 and 1.58. Drawing the line y = 2 - 2x on the same graph solves 4x cubed + 2x squared - 12x - 3 = 0, giving x is about -1.85, -0.25 and 1.63.

Why this method works

A graph solves an equation by turning its solutions into readable positions. Setting the curve to zero means y = 0, and every point with y = 0 lies on the x-axis, so the roots are just the crossings. For a related equation, the trick is to reuse the curve you already drew: by isolating the cubic block that both equations contain and substituting it, the awkward cubic terms cancel and leave a linear equation. That line's intersections with the curve are the only x-values where both the curve equation and the new equation hold at once, so they are exactly the solutions you want, found without any algebraic root-finding.

✓

Test x = -0.5 in the curve: 5 + 10(-0.5) - 2(0.25) - 4(-0.125) = 5 - 5 - 0.5 + 0.5 = 0, so x = -0.5 is a genuine root of the curve as read off the graph.