3 years ago

Give a # that makes 10 - x < 5 true.

1 answer:

5 0

It could be any number that is 4 and under

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Rufina [12.5K]

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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2 years ago