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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)

Screen Shot 2020 10 13 at 6.41.54 PM

(ii)

Screen Shot 2020 10 13 at 6.42.06 PM

(iii)

Screen Shot 2020 10 13 at 6.42.16 PM

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)

Screen Shot 2020 10 13 at 6.55.50 PM

ii. y = 8x2 – 2x

When y = 0

i.e.  8x2 – 2x = 0

Screen Shot 2020 10 13 at 7.01.17 PM

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

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