Higher Engineering Mathematics, Sixth Edition

Functions and their curves 197

18.9 Worked problems on curve sketching

Problem 13. Sketch the graphs of

(a)y= 2 x^2 + 12 x+ 20

(b)y=− 3 x^2 + 12 x− 15

(a) y= 2 x^2 + 12 x+20 is a parabola since the equa-
tion is a quadratic. To determine the turning
point:

Gradient=

dy

dx

= 4 x+ 12 =0 for a turning point.

Hence 4x=−12 andx=−3.

Whenx=−3,y= 2 (− 3 )^2 + 12 (− 3 )+ 20 =2.

Hence (−3, 2) are the co-ordinates of the turning

point

d^2 y

dx^2

=4, which is positive, hence (−3, 2) is a

minimum point.

Whenx=0,y=20, hence the curve cuts the

y-axis aty=20.

Thus knowing the curve passes through (−3, 2)

and (0, 20) and appreciating the general shape

of a parabola results in the sketch given in

Fig. 18.36.

(b) y=− 3 x^2 + 12 x−15 is also a parabola (but
‘upside down’ due to the minus sign in front of
thex^2 term).

Gradient=

dy

dx

=− 6 x+ 12 =0 for a turningpoint.

Hence 6x=12 andx=2.

Whenx=2,y=− 3 ( 2 )^2 + 12 ( 2 )− 15 =−3.

Hence (2,−3) are the co-ordinates of the turning

point

d^2 y

dx^2

=−6, which is negative, hence (2,−3) is a

maximum point.

Whenx=0,y=−15,hencethecurvecutstheaxis

aty=−15.

The curve is shown sketched in Fig. 18.36.

y 52 x^2112 x 120

y 523 x^2112 x 215

25

23

24 23 22 211023

5

210

10

20

215

225

x

y

2

220

15

Figure 18.36

Problem 14. Sketch the curves depicting the

following equations:

(a)x=

√

9 −y^2 (b)y^2 = 16 x

(c)xy= 5

(a) Squaring both sides of the equation and trans-

posing gives x^2 +y^2 =9. Comparing this with

the standard equation of a circle, centre ori-

gin and radiusa,i.e.x^2 +y^2 =a^2 ,showsthat

x^2 +y^2 =9 represents a circle, centre origin and

radius 3. A sketch of this circle is shown in

Fig. 18.37(a).

(b) The equationy^2 = 16 xis symmetrical about the

x-axis and having its vertex at the origin (0, 0).

Also, when x=1, y=±4. A sketch of this

parabola is shown in Fig. 18.37(b).

(c) The equation y=

a

x

represents a rectangular

hyperbola lying entirely within the first and third

quadrants. Transposingxy=5givesy=

5

x

,and

therefore represents the rectangular hyperbola

shown in Fig. 18.37(c).