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A function
has horizontal asymptote of a vertical asymptote of and an -intercept atPart (a): Let
be of the formPart (b): Let
be of the formPlease don't just give an answer, if you can, please explain.
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hi Enshrouded_
part a:
For a vertical asymptote the curve will become more and more vertical as x approaches the key value; in this case 3.
To make that happen set x + c = 0. Then as x approaches 3, x + c approaches zero and division by zero results in an infinite answer. So 3 + c = 0
For a horizontal asymptote, you want the function to tend to a fixed value as x approaches + or - infinity.
When f(x) = (ax+b)/(x+c) and x is approaching infinity the b and c values are insignificant and the function is the same as f(x) = (ax)/x = a
So a horizontal asymptote of y = -4 means we want a = -4
To find 'b', use the known point on the curve.
When you have an answer you can try it out at http://www.mathsisfun.com/data/function-grapher-old.php
part b.
When x = 3 you want 2x+t to be zero.
As x tends to infinity the function behaves like r/2 which will fix r. And finally find s by using the known point.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I don't get how you did the as x approaches 3 you get an infinite answer. Thanks.
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What I'm looking for is a denominator that is zero at x=3
So x + c = 0 => c = -3
Here is a demo with 1/(x-3)
You can see that as x gets closer to 3 from below or above, the value of the expression gets bigger and bigger. If you plot the points the curve becomes more and more vertical as you get closer to x = 3
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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