Question

determine the vertical asymptotes of the following function.
$f(x) = \frac{x^2 - 36}{x^2 - 13x + 42}$
the field below accepts a list of numbers or formulas separated by semicolons (e.g. $2;4;6$ or $x + 1;x - 1$).
enter each item from lowest to highest.

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 36\) as \((x - 6)(x + 6)\) (difference of squares: \(a^2 - b^2=(a - b)(a + b)\) with \(a = x\), \(b = 6\)).
Factor \(x^2 - 13x + 42\): find two numbers that multiply to \(42\) and add to \(-13\), which are \(-6\) and \(-7\). So, \(x^2 - 13x + 42=(x - 6)(x - 7)\).
Thus, \(f(x)=\frac{(x - 6)(x + 6)}{(x - 6)(x - 7)}\).

Step2: Cancel common factors

Cancel the common factor \((x - 6)\) (for \(x eq6\), since the original function is undefined at \(x = 6\) but it's a removable discontinuity, not a vertical asymptote).

Step3: Find vertical asymptote

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is not zero there).
Set the simplified denominator \(x - 7 = 0\), so \(x = 7\). Wait, but also check the original denominator's roots (excluding the canceled one). The original denominator roots are \(x = 6\) and \(x = 7\). But \(x = 6\) is canceled, so only \(x = 7\)? Wait, no—wait, let's re - check. Wait, when we factor, the original function is \(\frac{(x - 6)(x + 6)}{(x - 6)(x - 7)}\). The domain excludes \(x = 6\) and \(x = 7\). After canceling \((x - 6)\), the function is \(\frac{x + 6}{x - 7}\) for \(x eq6\). So, the vertical asymptote is where the simplified denominator is zero, i.e., \(x - 7 = 0\Rightarrow x = 7\)? Wait, no, wait—maybe I made a mistake. Wait, let's do it again.

Wait, vertical asymptotes are found by setting the denominator of the original function equal to zero, then excluding the roots that make the numerator zero (since those are holes).

Original denominator: \(x^2-13x + 42=(x - 6)(x - 7)\), roots at \(x = 6\) and \(x = 7\).

Original numerator: \(x^2 - 36=(x - 6)(x + 6)\), root at \(x = 6\) and \(x=-6\).

So, at \(x = 6\), both numerator and denominator are zero (hole). At \(x = 7\), denominator is zero and numerator is \(7^2-36=49 - 36 = 13 eq0\), so \(x = 7\) is a vertical asymptote. At \(x=-6\), numerator is zero, denominator is \((-6)^2-13*(-6)+42=36 + 78+42=156 eq0\), so \(x=-6\) is not a vertical asymptote. Wait, but wait, maybe I messed up the factoring? Wait, no, let's check the denominator again: \(x^2-13x + 42\). Let's use quadratic formula: \(x=\frac{13\pm\sqrt{169 - 168}}{2}=\frac{13\pm1}{2}\), so \(x=\frac{13 + 1}{2}=7\), \(x=\frac{13 - 1}{2}=6\). Correct. Numerator: \(x^2-36=(x - 6)(x + 6)\), correct. So, when \(x = 6\), both numerator and denominator are zero (hole), when \(x = 7\), denominator is zero, numerator is non - zero, so vertical asymptote at \(x = 7\)? Wait, but that seems wrong. Wait, no, wait the function is \(\frac{(x - 6)(x + 6)}{(x - 6)(x - 7)}\). If we cancel \((x - 6)\), we get \(\frac{x + 6}{x - 7}\) for \(x eq6\). So the graph of \(f(x)\) is the graph of \(\frac{x + 6}{x - 7}\) with a hole at \(x = 6\). The vertical asymptote of \(\frac{x + 6}{x - 7}\) is at \(x = 7\), since the denominator is zero there and numerator is \(7 + 6=13 eq0\). So the vertical asymptote is \(x = 7\)? Wait, but maybe I made a mistake in the numerator and denominator. Wait, let's plug \(x = 7\) into original function: denominator is \(49-91 + 42=0\), numerator is \(49-36 = 13 eq0\), so yes, vertical asymptote at \(x = 7\). Wait, but what about \(x=-6\)? Plug \(x=-6\) into denominator: \(36+78 + 42=156 eq0\), so \(x=-6\) is not a vertical asymptote. Wait, but the problem says "vertical asymptotes" (plural). Did I factor wrong? Wait, no, let's re - examine the function: \(f(x)=\frac{x^2 - 36}{x^2-13x + 42}\). Wait, maybe I factored the denominator wrong. Wait, \(x^2-13x + 42\): \(13\) is the sum, \(42\) is the product. Factors of \(42\): \(6\) and \(7\), since \(6 + 7=13\), but since the middle term is \(-13x\) and the constant term is \(+42\), both factors are negative: \((x - 6)(x - 7)\), correct. Numerator: \(x^2-36=(x - 6)(x + 6)\), correct. So the only vertical asymptote is \(x = 7\)? Wait, that seems odd. Wait, no, maybe I made a mistake in the cancellation. Wait, the original function is undefined at \(x = 6\) and \(x = 7\). At \(x = 6\), it's a hole (removable discontinuity) because both numerator and denominator are zero there. At \(x = 7\), denominator is zero and numerator is non - zero, so it's a vertical asymptote. So the vertical asymptote is \(x = 7\). Wait, but let's graph the function \(\frac{x + 6}{x - 7}\) (with a hole at \(x = 6\)). The vertical asymptote of \(\frac{x + 6}{x - 7}\) is at \(x = 7\), since as \(x\) approaches \(7\) from the left, \(x-7\) approaches \(0\) from the negative side, and \(x + 6\) approaches \(13\), so the function approaches \(-\infty\). As \(x\) approaches \(7\) from the right, \(x - 7\) approaches \(0\) from the positive side, and \(x+6\) approaches \(13\), so the function approaches \(+\infty\). So yes, vertical asymptote at \(x = 7\).

Answer:

\(7\)