So close yet so far

Algebra Level 3

f ( x ) = 4 x 2 − 10 x + 7 2 x − 5 f(x) = \dfrac{4x^2 -10x + 7}{2x - 5}

Consider the function above, what is the slope of the graph's oblique asymptote?


The answer is 2.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

f ( x ) = 4 x 2 − 10 x + 7 2 x − 5 = 2 x + 7 2 x − 5 f(x) = \dfrac{4x^2 -10x + 7}{2x - 5} = 2x + \dfrac{7}{2x-5}

By taking a polynomial division, we will get 2 x 2x as the quotient, and if y = f ( x ) = 2 x y = f(x) = 2x , then 0 = 7 2 x − 5 0 = \dfrac{7}{2x-5} , where the graphs only approach each other at plus or minus infinity.

Hence, y = 2 x y = 2x is the graph's oblique asymptote, resulting in a slope of 2 .

Below is the graph of the function f ( x ) = 4 x 2 − 10 x + 7 2 x − 5 f(x) = \dfrac{4x^2 -10x + 7}{2x - 5} (red) with its oblique asymptote y = f ( x ) = 2 x y = f(x) = 2x (blue):

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