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jaredroy23

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Two statements, P and Q, about f(x) appear just below. Examine these statements and then choose the answer which best describes them.
f(x) = [g(x)]/[h(x)] where g(x) = x^4 + 2x^2 + 1 and h(x) = x^2 + 1 .

P: The graph of f(x) is symmetric with respect to the y-axis.

Q: The graph of f(x) has an oblique asymptote.

a. Only P is true.
b. Only Q is true.
c. Both P and Q are true.
d. Both P and Q are false.
 
Hello, jaredroy23!

Two statements, P and Q, about f(x) appear just below.
Examine these statements and then choose the answer which best describes them.
\(\displaystyle f(x)\,=\,\frac{g(x)}{h(x)}\), where \(\displaystyle g(x)\,=\,x^4\,+\,2x^2\,+\,1\) and \(\displaystyle h(x)\,=\,x^2\,+\,1\)

P: The graph of f(x) is symmetric with respect to the y-axis.

Q: The graph of f(x) has an oblique asymptote.

a. Only P is true . b. Only Q is true . c. Both P and Q are true . d. Both P and Q are false
Click to expand...
\(\displaystyle \L f(x)\;=\;\frac{g(x)}{h(x)}\;=\;\frac{x^4\,+\,2x^2\,+\,1}{x^2\,+\,1}\;=\;\frac{(x^2\,+\,1)^2}{x^2\,+\,1}\;=\;x^2\,+\,1\)

. . We can cancel because \(\displaystyle x^2\,+\,1\:\neq\:0\)


\(\displaystyle f(x)\) is a parabola with vertex \(\displaystyle (0,1)\).

It is symmetric to the y-axis, but has no asymptotes.

Answer: (a) only P is true.
 
So what have you tried? What are your thoughts? What is your specific question on this exercise? You've formed the "f(x)" fraction, you've drawn the graph, you've done the long division or whatever, and... then what?

(For a related discussion of an almost-identical exercise, try this thread.)

Thank you.

Eliz.
 
stapel said:
So what have you tried? What are your thoughts? What is your specific question on this exercise?
Click to expand...
Never mind.

Eliz.
 
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