Back to Course

0% Complete
0/0 Steps

• ## Follow us

Lesson 7, Topic 2
In Progress

# Finding An Equation Of A Curve From Its Graph

Lesson Progress
0% Complete

### Introduction:

The equation of a curve represented by a sketch of the curve can be obtained if we know where the curve crosses the axes.

### Curve Sketching:

Example 1

The sketches below are of the form y = ax2 + bx + c, find the values of a, b and c respectively.

(i)

(ii)

(iii)

Solution:

(i) When y = 0, x = -1 or x = $$\frac{5}{2}$$

i.e. $$\left( \scriptsize x + 1 \right) \left( \scriptsize x \; – \; \frac{5}{2}\right) \scriptsize = 0$$

i.e. $$\scriptsize x^2 \; – \; \normalsize \frac{5x}{2} \;+ \: \scriptsize x \; – \; \normalsize \frac{5x}{2} \scriptsize = 0$$

Multiply both sides by 2

i.e. $$\scriptsize 2x^2 – 5x + 2x – 5 = 0$$

i.e. $$\scriptsize 2x^2 – 3x – 5 = 0$$

when x = 0, y = -5

Also y -intercept = -5

∴    the required equation is $$\scriptsize y = 2x^2 – 3x – 5$$

i.e.  a = 2, b = -3 and c = -5

ii. When y = 0, x = -1 or x = 4

i.e. (x+1)(x-4) = 0

i.e. $$\scriptsize x^2 – 4x + x – 4 = 0$$

i.e. x2 – 3x – 4 = 0

Since y-intercept = 4

∴y= x2 -3x – 4     …….required equation

i.e. a = 1, b = -3 and c = -4.

(iii) When y = 0, x = -3 or x = 3

i.e. x + 3 = 0 or x – 3 = 0

i.e. (x + 3)(x – 3) = 0

i.e. x2 – 3x + 3x – 9 = 0

i.e. x2 – 9 = 0

y-intercept = -9

i.e. a = 1, b = 0 and c = -9

Example 2:

Sketch the graph of each of the following:

(i) y = (2 + x)(3 – x) (ii) y = 8x2 – 2x

Solution:

i. When y = 0

i.e. (2 + x)(3 – x) = 0

i.e.  x + 2 = 0 or 3 – x =

i.e. x = -2 or x = 3

Therefore, the curve cuts the x-axis at x = 3 and -2

When x = 0

y = (2 + x)(3 – x)

y = 6 + x – x2

i.e. y = 6   (y-intercept)

ii. y = 8x2 – 2x

When y = 0

i.e.  8x2 – 2x = 0

i.e. 2x (4x – 1) = 0

i.e. 2x = 0  or 4x -1 = 0

i.e. x = 0 or x = $$\frac{1}{4}$$

when x = 0, y = 0

error: