QUESTION IMAGE
Question
find the equation of the slant asymptote of the rational function given below.
$f(x) = \frac{x^2 - 11x + 48}{-4 + x}$
Step1: Simplify the denominator
Rewrite the denominator as \(x - 4\) since \(-4 + x=x - 4\). So the function is \(f(x)=\frac{x^{2}-11x + 48}{x - 4}\).
Step2: Perform polynomial long division
Divide the numerator \(x^{2}-11x + 48\) by the denominator \(x - 4\).
- Divide the leading term of the numerator \(x^{2}\) by the leading term of the denominator \(x\), we get \(x\). Multiply \(x-4\) by \(x\) to get \(x^{2}-4x\).
- Subtract \(x^{2}-4x\) from \(x^{2}-11x + 48\): \((x^{2}-11x + 48)-(x^{2}-4x)=x^{2}-11x + 48 - x^{2}+4x=-7x + 48\).
- Now divide the leading term of \(-7x\) by \(x\) (the leading term of the denominator), we get \(-7\). Multiply \(x - 4\) by \(-7\) to get \(-7x+28\).
- Subtract \(-7x + 28\) from \(-7x + 48\): \((-7x + 48)-(-7x + 28)=-7x + 48+7x - 28 = 20\).
- So, \(\frac{x^{2}-11x + 48}{x - 4}=x-7+\frac{20}{x - 4}\).
Step3: Determine the slant asymptote
As \(x\) approaches \(\pm\infty\), the term \(\frac{20}{x - 4}\) approaches \(0\). So the slant asymptote is the line \(y=x - 7\).
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The equation of the slant asymptote is \(y = x-7\)