Use the guidelines to sketch the curve. y = 3x^2/x^2 - 16 The domain is {x | x^2

Question

Use the guidelines to sketch the curve. y = 3x^2/x^2 - 16 The domain is {x | x^2 - 16 = 0} = {x | x plusminus4} = (-infinity,) u () U (infinity) The x- and y-intercepts are both 0. Since f(-x) = f(x), the function is. The curve is symmetric about the y-axis. Therefore the line y = is a horizontal asymptote. Since the denominator is 0 when x = plusminus 4, we compute the following limits: Therefore the lines x = -4 and x = are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the top figure. The only critical number is x =. Since f' changes from positive to negative at 0, f(0) = 0 is a local maximum by the First Derivative Test. Since 288x^2 + 1536 > 0 for all x, we have f "(x) >0 x^2 - 16 >0 |x| > and f "(x)

Explanation / Answer

I am allowed to answer only four parts at a time.

A)
x != 4 and x!=-4
So, domain is:
(-inf,-4) U (-4,4) U (4,inf)

B)
x intercept = 0
y intercept =0

C)
Function is EVEN

D)
Y=3 IS HORIZONTAL ASYMPTODES
limit on left will be infinity and that on right will be - infinity
x=-4 and x=4 are vertical asymptodes

E)
f is increasing on (-inf,-4)) and (4,0)
f is decreasing at (0,4) and (4,inf)

F)
X= 0

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