Higher Engineering Mathematics

210 GRAPHS

19.9 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. 19.36.

(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 turning

point.

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, hence the curve cuts the

axis aty=−15.

The curve is shown sketched in Fig. 19.36.

− 4 − 3 − 2 − 1 0123 x

5

2

10

20

y

15

− 5

− 10

− 15

− 20

− 25

y = − 3 x^2 + 12 x− 15

y = 2 x^2 + 12 x+ 20

− 3

Figure 19.36

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 givesx^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 , shows that

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

radius 3. A sketch of this circle is shown in

Fig. 19.37(a).

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

thex-axis and having its vertex at the origin

(0, 0). Also, whenx=1,y=±4. A sketch of

this parabola is shown in Fig. 19.37(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. 19.37(c).