Cubic Graph and Straight Line Solution Interactive

Explore tangent at different x-values: x = -1.0, gradient ≈ 2.20

curve plotted points tangent straight line solution

Question

y = ¹⁄₅x³ − ⁴⁄₅x²

Draw the graph for −2 ≤ x ≤ 5, estimate the tangent gradient at (−1, −1), then use the graph to solve:

x³ − 4x² + 5x − 10 = 0

Consistent equation style:
When writing a polynomial in x, place the highest power first, then descend. For example, write y = −x + 2, not y = 2 − x.

Table of values

x	−2	−1	0	1	2	3	4	5
y	−4.8	−1	0	−0.6	−1.6	−1.8	0	5

Teaching steps

(a) Draw the curve:
Plot the table points and join them with a smooth curve.

(b) Tangent gradient:
Draw a tangent at (−1, −1). Choose two convenient points on the tangent, then use:
gradient = change in y ÷ change in x

(c)(i) Suitable straight line:
Rewrite the cubic equation so that it uses the curve already drawn. The straight line is written as y = −x + 2.

(c)(ii) Solve by intersection:
The solution is the x-coordinate where the curve meets the straight line.

Answers visualised

(b) Tangent gradient using two points
Formula:
gradient = ^{change in y}⁄_{change in x} = ^{y₂ − y₁}⁄_{x₂ − x₁}

From the tangent at (−1, −1), take two convenient points on the tangent:
(x₁, y₁) = (−1, −1), (x₂, y₂) = (0, 1.2)

Substitute:
gradient = ^{1.2 − (−1)}⁄_{0 − (−1)} = ^2.2⁄₁ = 2.2

Check using differentiation:
y = ¹⁄₅x³ − ⁴⁄₅x²
dy/dx = ³⁄₅x² − ⁸⁄₅x
At x = −1:
gradient = ³⁄₅(−1)² − ⁸⁄₅(−1) = ³⁄₅ + ⁸⁄₅ = ¹¹⁄₅ = 2.2

(c)(i) Straight line:
x³ − 4x² + 5x − 10 = 0
Divide by 5:
¹⁄₅x³ − ⁴⁄₅x² + x − 2 = 0
Since y = ¹⁄₅x³ − ⁴⁄₅x² ,
y + x − 2 = 0
Rearrange:
y = −x + 2
Therefore, the suitable straight line is:
y = −x + 2

(c)(ii) Solution:
The curve intersects y = −x + 2 at about:
x ≈ 3.39
So the graph answer is usually:
x ≈ 3.4

Important teaching point: the cubic equation is not solved by looking where the blue curve crosses the x-axis. It is solved by finding where the blue curve meets the green straight line y = −x + 2.