Question

please do this question as it is very urgwnt for me and for my assignments.



Trace the curve y = 8/(4-x^2) and state all the properties you use to trace it.

Answer 1

Solution;

All the properties used to trace the graph are;

1)Domain.

Find the domain of the function and determine points of discontinuity.

$f(x)=\frac{8}{4-x^2}$

Rewrite as follows;

$f(x)=\frac{8}{-(x^2-4)}=\frac{8}{-(x+2)(x-2)}$

For domain gaps,take the denominator to be equal to zero;

-(x+2)(x-2)=0

Therefore,the points of discontinuity are;

$x=2 or -2$

2) Intercepts.

For y -intercept,take x=0

$f(0)=\frac{8}{4-0}=2$

y-intercept is (0,2)

The curve has no x-intercept.

3) Symmetry of the function.

The curve has neither axially symmetry nor point symmetry.

4)Intervals of increase or decrease.

5)Local maximum and minimum.

For (4) and (5),

$f(x)=\frac{8}{4-x^2}$

The first derivative of f(x);

$f'(x)=\frac{16x}{(-1)^2(x+2)^2(x-2)^2}$

The second derivative is;

$f''(x)=\frac{48x^4-128x^2-256}{-(x+2)^4(x-2)^4}$

Looking for turning points;

Equate denominator of f'(x) to 0;

$(-1)^2(x+2)^2(x-2)^2=0$

Hence the gaps are at;

x=2 or -2

If f'(x)=0;

$\frac{16x}{(-1)^2(x+2)^2(x-2)^2}=0$

Multiply both sides with the denominator;

16x=0

x=0

Turning points could be at x=0.

Substitute 0 into f''(x);

$f''(0)=\frac{48(0)-128(0)-256}{-(0+2)^4(0-2)^4}$ =1

Check if f'(x) changes sign to confirm;

Use -1 and 1;

$f'(1)=\frac{16}{(-1)^2(1+2)^2(1-2)^2}=1.78$

$f'(-1)=\frac{16×-1}{(-1)^2(-1+2)^2(-1-2)^2}=-1.78$

There is change from positive to negative,hence there exist a turning point;

Substitute 0 into the f(x);

$f(0)=\frac{8}{4-0}=0$

There is a minimum turning point at ;

(0,2)