Let's start with 2 vertical asymptotes x=3 and x= -3
That means that the denominator must contain (x-3)(x+3)
Since it passes through (0,0), it would suggest that we have a factor of ax at the top
If the x-axis is a horizontal asymptote, then the numerator must be a degree lower than the
degree of the denominator, confirming my suggestion in the last sentence
We have the point (4, 8/9) to worry about
So how about starting with y = ax/((x-3)(x+3))
or y = ax/(x^2 - 9)
testing the point (0,0), yup that works
if x = 4, y = 8/9
8/9 = 4a/(16-9)
56 = 36a
a = 56/36 = 14/9
the function f(x) = 14x/(9(x^2 - 9)) or 14x/(9x^2 - 81) satisfies all your stated conditions
check:
https://www.wolframalpha.com/input/?i=graph+f%28x%29+%3D+14x%2F%289%28x%5E2+-+9%29%29+from+-8+to+8
YUP!!
Write the equation of the function that passes through the points (0,0) and (4, 8/9 (fraction) ); has the x-axis as a horizontal asymptote; and has 2 vertical asymptotes x=3 and x= -3. Show calculations.
I’m genuinely confused on how to do this question and tried everything :(
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1 answer