f(x) = $\frac{x^{2}-4}{x^{2} -25}$
f(-x) = $\frac{(-x)^{2}-4}{(-x)^{2} -25}$
= $\frac{x^{2}-4}{x^{2} -25}$
f(-x) = f(x) so the function is even.
Step III : Find x and y intercepts
To find x-intercept , plug in y=0 in the given function
f(x) = $\frac{x^{2}-4}{x^{2} -25}$
0 = $\frac{x^{2}-4}{x^{2} -25}$
$x^{2}$-4 = 0
$x^{2}$ = 4
x = $\pm 2$
X-intercept : (-2,0) and (2,0)
To find the y-intercept, plug in x = 0 in the given function
f(x) = $\frac{x^{2}-4}{x^{2} -25}$
f(0) = $\frac{0^{2}-4}{0^{2} -25}$
y = $\frac{-4}{-25}$
y = 4/25
Y-intercept : (0,4/25)
Step IV : Find vertical asymptotes
For vertical asymptotes, set the denominator equal to zero
$x^{2}$ - 25 = 0
$x^{2}$ = 25
x = $\pm$ 5
Vertical asymptote : x = 5 and x = -5
Step V : Find horizontal asymptotes
As the degree of the numerator is equal to the degree of denominator then the horizontal asymptote is
y = $\frac{Leading coefficient of numerator}{Leading coefficient of denominator} $
so Horizontal asymptote : y = 1
Step IV : Find the derivatives for( differentiability, Relative extrema, Concavity, Points of inflection)
f(x) = $\frac{x^{2}-4}{x^{2} -25}$
=$\frac{(x^{2} -25)\frac{ \text{d}}{\text{d}x}(x^{2} -4)-(x^{2} -4)\frac{\text{d}}{\text{d}x}(x^{2}-25)}{(x^{2}-25)^{2}}$
=$=\frac{(x^{2} -25)(2x)-(x^{2} -4)(2x)}{(x^{2}-25)^{2}}$
f '(x) = $\frac{-42x}{(x^{2}- 25)^{2}}$
f "(x) = $ \frac{42(3x^{2}+25) }{(x^{2}- 25)^{3}}$
For critical points f '(x) = 0
$\frac{-42x}{(x^{2}- 25)^{2}}$ = 0
-42x = 0
x= 0
f '(x) is undefined at x = $\pm$ 5
Interval for critical points : ($\infty$, -5) (-5,0)(0,5) (5,$\infty$ )
Set the table :
Point of inflection f "(x) $\neq$ 0
(-∞,-5)
x=-5
(-5,0)
x=0
(0,5)
x=5
(5,∞)
f(x)
undf
4/25
undf
f '(x)
+
undf
+
0
-
undf
-
f "(x)
+
undf
-
-
-
undf
+
Conclusion
increasing
Increasing
decreasing
decreasing
concave up
concave down
concave down
concave up
Rel.max