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consider the following function. f(x)=\\frac{x^{2}-15x + 56}{x^{2}-17x …

Question

consider the following function.

f(x)=\frac{x^{2}-15x + 56}{x^{2}-17x + 72}

a) determine the vertical asymptote(s). enter each asymptote as a separate equation,
separating multiple asymptotes with a comma. if there are no vertical asymptotes, enter
none.

b) determine the x - value(s) of any removable discontinuities. separate multiple
removable discontinuities with a comma. if there are no removable discontinuities, enter
none.

Explanation:

Step1: Factor numerator and denominator

For numerator \(x^{2}-15x + 56\), using \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factoring by finding two numbers that multiply to \(56\) and add to \(-15\). We get \(x^{2}-15x + 56=(x - 7)(x - 8)\).
For denominator \(x^{2}-17x + 72\), find two numbers that multiply to \(72\) and add to \(-17\). We get \(x^{2}-17x + 72=(x - 8)(x - 9)\).
So \(f(x)=\frac{(x - 7)(x - 8)}{(x - 8)(x - 9)}\).

Step2: Simplify the function

Cancel out the common factor \((x - 8)\) (for \(x
eq8\)). The simplified function is \(f(x)=\frac{x - 7}{x - 9}\), \(x
eq8\).

Step3: Find vertical asymptote

Vertical asymptote occurs when the denominator of the simplified function (after canceling non - zero common factors) is zero. Set \(x-9 = 0\), so \(x = 9\).

Step4: Find removable discontinuity

Removable discontinuity occurs at the \(x\) - value that was canceled out. Here \(x = 8\).

Answer:

a) \(x = 9\)
b) \(8\)