Question

identify the vertical asymptote(s) of the function. choose two correct answers.\\( f(x) = \frac{x + 2}{x^2 - 3x - 4} \\)\\( x = -4 \\) \\( x = 1 \\) \\( x = -1 \\) \\( x = -2 \\) \\( x = 4 \\) \\( x = 2 \\)

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the denominator of the rational function. The denominator is \(x^2 - 3x - 4\). We factor it as follows:
We need two numbers that multiply to \(-4\) and add to \(-3\). The numbers are \(-4\) and \(1\). So, \(x^2 - 3x - 4=(x - 4)(x + 1)\)? Wait, no, wait: \((x - 4)(x + 1)=x^2 - 3x - 4\)? Let's check: \(x\times x=x^2\), \(x\times1 = x\), \(-4\times x=-4x\), \(-4\times1=-4\). Then \(x^2 + x - 4x - 4=x^2 - 3x - 4\). Yes, correct. Wait, but wait, the numerator is \(x + 2\), which doesn't share a common factor with the denominator (since \(x + 2\) and \((x - 4)(x + 1)\) have no common linear factors).

Step2: Find values that make denominator zero

Vertical asymptotes occur where the denominator is zero (and numerator is not zero, to avoid a hole). So we set the denominator equal to zero:
\((x - 4)(x + 1)=0\)
Using the zero - product property, we have \(x - 4 = 0\) or \(x+1 = 0\).
Solving \(x - 4 = 0\) gives \(x = 4\).
Solving \(x + 1 = 0\) gives \(x=-1\). Wait, but wait, earlier I made a mistake in factoring? Wait, let's re - factor the denominator \(x^2-3x - 4\) again. The quadratic formula: for \(ax^2+bx + c\), \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b=-3\), \(c=-4\). Then \(x=\frac{3\pm\sqrt{9+16}}{2}=\frac{3\pm\sqrt{25}}{2}=\frac{3\pm5}{2}\). So \(x=\frac{3 + 5}{2}=4\) and \(x=\frac{3-5}{2}=-1\). So the denominator factors as \((x - 4)(x + 1)\). Wait, but the options have \(x=-4\), \(x = 1\), etc. Wait, did I factor wrong? Wait, maybe I made a mistake in the sign. Let's check again: \(x^2-3x - 4\). Let's try \((x + 4)(x - 1)\): \(x^2 - x+4x - 4=x^2 + 3x - 4\), no. \((x - 4)(x + 1)=x^2 - 3x - 4\), correct. Wait, but the options include \(x=-1\) and \(x = 4\). Let's check the original function: \(f(x)=\frac{x + 2}{x^2-3x - 4}\). So when \(x=-1\), denominator is \((-1)^2-3\times(-1)-4=1 + 3-4 = 0\), numerator is \(-1 + 2=1
eq0\), so \(x=-1\) is a vertical asymptote. When \(x = 4\), denominator is \(16-12 - 4=0\), numerator is \(4 + 2=6
eq0\), so \(x = 4\) is a vertical asymptote. Wait, but the options have \(x=-1\) and \(x = 4\) as options? Wait, the options are \(x=-4\), \(x = 1\), \(x=-1\), \(x=-2\), \(x = 4\), \(x = 2\). So the correct ones are \(x=-1\) and \(x = 4\).

Wait, but let's re - check the factoring. Maybe I had a sign error. Let's do the factoring again. \(x^2-3x - 4\). We need two numbers \(m\) and \(n\) such that \(m\times n=-4\) and \(m + n=-3\). Let's list the factor pairs of \(-4\): \((-4,1)\), \((4,-1)\), \((-2,2)\). For \((-4,1)\): \(-4+1=-3\), which is correct. So the factors are \((x - 4)(x + 1)\) because \(x\times x=x^2\), \(x\times1=x\), \(-4\times x=-4x\), \(-4\times1=-4\), and \(x-4x=-3x\). So that's correct. So the denominator is zero when \(x = 4\) (from \(x - 4 = 0\)) and \(x=-1\) (from \(x + 1 = 0\)). So the vertical asymptotes are \(x=-1\) and \(x = 4\).

Answer:

The two correct answers are \(x=-1\) and \(x = 4\), so the options are:

• \(x=-1\)
• \(x = 4\)