Posted by Gabby on Sunday, April 28, 2013 at 4:19pm.
I will assume you meant:
f(x) = (x^2 + 1)/(x^2 - 9)
= (x^2 + 1)/((x-3)(x+3))
a) for y intercepts , let x = 0
then y = 1/-9 = - 1/9
for x intercepts, let y = 0
(x^2 + 1)/(x^2 - 9) = 0
then x^2 + 1 = 0
x^2 = -1 , which has no real solutions
so there are no x-intercepts
b) using the quotient rule
dy/dx = ( (x^2-9)(2x) - (x^2+1)(2x))/(x^-9)^2
= -20x/(x^2 - 9)^2
c) --- critical values are found in many of the other parts.
d) for local extrema, dy/dx = 0
-20x/(x^2 - 9)^2 = 0
-20x = 0
x = 0 , then f(0) = -1/9
time to look at the graph using Wolfram, ( my other answers are confirmed here)
http://www.wolframalpha.com/input/?i=y+%3D+%28x%5E2+%2B+1%29%2F%28x%5E2+-+9%29
there is a local maximum at -1/9 , when x = 0
e) y '' = ((x^2 - 9)^2 (-20) + 20x(2)(x^2 - 9)(2x) )/(x^2 - 9)^4
= 60(x^2 + 3)/(x^2 - 9)^3
f) concave up or concave down depends on y''
if y'' is > 0 for a given x, then the curve is concave up at that value
if y'' is < 0 for .... concave down
While the numerator of y’’ is always positive,
the denominator of y’’ could be + or –
for -3 < x < 3 , the value of (x^2 – 9)^3 is negative, all all values yield a positive result.
So the curve is concave down for -3 < x < 3
For all other values it is concave up
Of course it is undefined for x = ± 3, giving us vertical asymptotes at
X = 3 and x = -3
g)
points of inflection occur when y'' = 0
60(x^2 + 3)/(x^2 - 9)^3 = 0
60(x^2 + 3) = 0
x^2 + 3 = 0
x^2 = -3
No solution, thus no point of inflection