How do you sketch the curve #y=x/(x^2-9)# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Answer 1

Local Maximum/Minimum

Start by finding the first derivative.

#y' = (1(x^2 - 9) - 2x(x))/(x^2 - 9)^2#
#y' = (x^2 - 9 - 2x^2)/(x^2 - 9)^2#
#y' = -(x^2 + 9)/(x^2 - 9)^2#
Critical numbers will occur whenever the derivative equals #0# or the derivative is undefined.
The derivative is undefined at #x = +- 3#, but this is undefined in the actual function as well, so these are actually not critical numbers. The derivative is never equal to #0# since #x^2 + 9 = 0# has no real solution.

Therefore, there are no critical numbers and thus no local max/min.

Points of inflection

Now find the second derivative. I used a derivative calculator for this, because it's very long to do by hand.

#y'' = (2x(x^2 + 27))/(x^2 - 9)^3#
This will have a point of inflection at #x= 0#. Let's now determine concavity.
Test Point: #x= -1#
#y''(-1) = (2(-1)((-1)^2 + 27))/((-1)^2 - 9)^3#
#y''(-1) = "positive"#
Therefore, #y# is concave up on #(-oo, 0)# and concave down on #(0, oo)#.

Asymptotes

This will have horizontal and vertical asymptotes.

Vertical

We factor the denominator:

#x^2 - 9 = (x + 3)(x - 3)#
This means there are vertical asymptotes at #x = -3# and #x = +3#.

Horizontal

#y = lim_(x->oo) (x/x^2)/(x^2/x^2 - 9/x^2)#
#y = lim_(x->oo) (1/x)/(1 - 9/x^2)#
#y = lim_(x->oo) (lim_(x->oo) (1/x))/((lim_(x->oo) (1) - lim_(x->oo) 9/x^2)#
By the identity #lim_(x-> oo) 1/x = 0#, we have:
#y = 0/(1 - 0)#
#y = 0#
Therefore, there will be a horizontal asymptote at #y= 0#.

Intercepts

We find the x-intercepts by setting #y# to #0#.
#0 = x/(x^2 - 9)#
#0 = x#
We find the y-intercepts by setting #x# to #0#.
#y = 0/(0^2 - 9)#
#y = 0#

We can now more or less trace the graph. Many people, including myself, like to prepare a table of values before graphing, but I'll leave that choice up to you.

graph{y = x/(x^2 - 9) [-28.87, 28.86, -14.43, 14.44]}

Hopefully this helps!

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Answer 2
  1. Find the intercepts:

    • y-intercept: Set (x = 0) to get (y = 0).
    • x-intercepts: Set (y = 0) to solve for (x).
  2. Find asymptotes:

    • Vertical asymptotes: Set the denominator (x^2 - 9) equal to zero and solve for (x). These are at (x = -3) and (x = 3).
    • Horizontal asymptote: Compare degrees of numerator and denominator. Since both are of degree 1, divide the leading coefficients: (y = \frac{1}{1}), so (y = 1) is the horizontal asymptote.
  3. Find critical points for extrema and inflection:

    • Find (f'(x)) and set it equal to zero to find critical points.
    • (f'(x) = \frac{-2x^2 + 9}{(x^2 - 9)^2}).
    • Critical points are at (x = -3/√2) and (x = 3/√2).
  4. Test for local extrema:

    • Use the second derivative test or first derivative test at the critical points.
    • Second derivative (f''(x) = \frac{-2(3x^2 - 13)}{(x^2 - 9)^3}).
    • Evaluate at the critical points to determine nature of extrema.
  5. Test for inflection points:

    • Set (f''(x) = 0) to find possible inflection points.
    • Solve (3x^2 - 13 = 0) to get (x = ±\sqrt{\frac{13}{3}}).
  6. Plot the points:

    • Plot intercepts, asymptotes, local max/min, and inflection points.
    • Sketch the curve using this information.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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