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- anonymous

How do you graph f(x)=(x+4)/(x-1)?

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- anonymous

How do you graph f(x)=(x+4)/(x-1)?

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- anonymous

f(x) is like y so you put in (x+4)/(x-1) in the graphing calculator

- anonymous

or it would be x-4 into the calculator because x/x = x and 4/-1 = -4

- anonymous

How do you graph it without a calculator though?

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- anonymous

-4 would be the y intercept and x would be the slope which is 1/1

- anonymous

The best way to graph these functions is to first find your intercepts and asymptotes. You find the x-intercept when you let y = 0, so the x-intercept is (-4, 0). You find the y-intercept when you let x = 0, so the y-intercept is (0, -4). Plot these two points on the xy-plane.
To find vertical asymptotes, set the denominator x - 1 = 0 -> x = 1. Draw a vertical dotted line at x = 1 to show that the graph CANNOT CROSS this line.
To find horizontal asymptotes, check the degrees (the highest exponent) of the numerator and the denominator. Since the degrees of the numerator and denominator are both 1, you must take a ratio of the coefficients of the leading terms (the terms in the numerator and denominator with the greatest exponents). This ratio is 1, so y = 1 is your horizontal asymptote. Draw a dotted line on y = 1.
Then, connect the dots (-4, 0) and (0, -4) with a nice smooth curve, but have it go past both points to "hug" the asymptotes without crossing them. Then, put another "wing" in the upper right box created by the asymptotes.

- anonymous

Thank you so much! That makes sense. I should probably ask this in a different question, but here it is anyways:
How would I graph 5/(x+1)(x-3)? Since there's no x in the numerator?

- anonymous

Think about it using the methods described above. For y-intercepts, let x = 0 to get 5/(1*-3) = -5/3... (0, -5/3) is your y-intercept. If you tried to find an x-intercept by letting y =0, you would just get a false statement 0 = 5, so there's no x-intercept.
The vertical asymptotes are where the denominator (x+1)(x - 3) = 0, so at x = -1 and at x = 3.
For horizontal asymptotes, if the degree of the denominator is greater than the degree of the numerator (in this case, the numerator has a degree of 0 and the denominator has a degree of 2), then the horizontal asymptote is at y = 0. This should somewhat make sense being that we don't have an x-intercept.
Because we're dealing with a couple asymptotes, we should get in the habit of checking intervals. What you do is pick an x value in each "yard" created by the asymptotes and see if the "wing" is positive or negative.
When we choose a value in the x < -1 yard, the y-value is positive, so we draw a wing above the x-axis following the asymptotes. When we choose a value in the -1 < x < 3 yard, the y-value is negative, so we draw a parabola-looking think below the x-axis following the asymptotes. When we choose a value in the x > 3 yard, the y-value is positive, so the wing is above the x-axis.

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