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hichkok12 [17]
3 years ago
12

Make a rational equations with the following requirements

Mathematics
1 answer:
Rufina [12.5K]3 years ago
4 0

Answer:

f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}

Step-by-step explanation:

Since f(x) has asymptotes at x = 5 and x = -5, then the denominator of the rational function contains the terms (x - 5) and (x + 5):

f(x)=\dfrac{?}{(x-5)(x+5)}

Since f(x) has x-intercepts at x = 2 and x = 1, then the numerator of the rational function contains the terms (x - 2) and (x - 1):

f(x)=\dfrac{A(x-1)(x-2)}{(x-5)(x+5)}

Now substitute the point (0, 4) and solve for A:

f(0)=4\\\\\implies \dfrac{A(0-1)(0-2)}{(0-5)(0+5)}=4\\\\\\\implies -\dfrac{2}{25}A=4\\\\\\\implies A=-50

So final rational function:

f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}

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