Question

question 5
determine the vertical asymptote(s) of the graph of the function.
$f(x) = \frac{x + 5}{7x^2 - 38x - 24}$
$\bigcirc$ $x = -6$ and $x = \frac{4}{7}$
$\bigcirc$ none
$\bigcirc$ $x = 6$ and $x = -\frac{4}{7}$
$\bigcirc$ $x = 5$

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the quadratic in the denominator \(7x^2 - 38x - 24\). We need two numbers that multiply to \(7\times(-24)= -168\) and add to \(-38\). The numbers are \(-42\) and \(4\). So we rewrite the middle term:

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Wait, no, wait. Wait, let's check again. Wait, maybe I made a mistake. Let's do it properly. The quadratic is \(7x^2-38x - 24\). Let's use the quadratic formula \(x=\frac{38\pm\sqrt{38^2 - 4\times7\times(-24)}}{2\times7}=\frac{38\pm\sqrt{1444 + 672}}{14}=\frac{38\pm\sqrt{2116}}{14}=\frac{38\pm46}{14}\). So the roots are \(\frac{38 + 46}{14}=\frac{84}{14}=6\) and \(\frac{38 - 46}{14}=\frac{-8}{14}=-\frac{4}{7}\). Wait, so the denominator factors as \(7(x - 6)(x+\frac{4}{7})\)? No, wait, the quadratic factors as \((7x + 4)(x - 6)\)? Wait, no, \((7x + 4)(x - 6)=7x^2-42x + 4x - 24=7x^2-38x - 24\). Yes, that's correct. So the denominator is \((7x + 4)(x - 6)\).

Step2: Find values that make denominator zero

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero there). So we set each factor equal to zero:

• For \(7x + 4 = 0\), we solve for \(x\): \(7x=-4\Rightarrow x = -\frac{4}{7}\)
• For \(x - 6 = 0\), we solve for \(x\): \(x = 6\)

Now we check the numerator at these \(x\)-values. The numerator is \(x + 5\). At \(x = 6\), numerator is \(6 + 5 = 11
eq0\). At \(x = -\frac{4}{7}\), numerator is \(-\frac{4}{7}+5=\frac{-4 + 35}{7}=\frac{31}{7}
eq0\). So the vertical asymptotes are at \(x = 6\) and \(x = -\frac{4}{7}\).

Answer:

\(x = 6\) and \(x = -\frac{4}{7}\) (corresponding to the option "x = 6 and x = \(-\frac{4}{7}\)")