Question

what are the vertical asymptotes of the function $f(x) = \frac{4x + 8}{x^2 + 3x - 4}$?\
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$\bigcirc$ $x = -1$ and $x = -2$\
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$\bigcirc$ $x = -1$ and $x = 2$\
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$\bigcirc$ $x = -1$ and $x = -4$\
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$\bigcirc$ $x = 1$ and $x = -4$

Explanation:

Step1: Factor denominator

Factor \(x^2 + 3x - 4\). We need two numbers that multiply to \(-4\) and add to \(3\). Those 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: \(x\times x=x^2\), \(x\times(-1)=-x\), \(4\times x = 4x\), \(4\times(-1)=-4\). Then \(-x + 4x=3x\), so yes, \(x^2 + 3x - 4=(x + 4)(x - 1)\)? Wait, but wait, the numerator is \(4x + 8 = 4(x + 2)\). Now, vertical asymptotes occur where the denominator is zero (and numerator is not zero at those points). So set denominator equal to zero: \((x + 4)(x - 1)=0\)? Wait, no, wait, maybe I factored wrong. Wait, let's re - factor \(x^2+3x - 4\). Wait, the quadratic formula: for \(ax^2+bx + c\), roots are \(\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 3\), \(c=-4\). So discriminant is \(9+16 = 25\), roots are \(\frac{-3\pm5}{2}\). So \(\frac{-3 + 5}{2}=\frac{2}{2}=1\), \(\frac{-3-5}{2}=\frac{-8}{2}=-4\). So the denominator factors as \((x - 1)(x + 4)\). Now, the numerator is \(4x + 8=4(x + 2)\). Now, we check if the numerator is zero at \(x = 1\) or \(x=-4\). At \(x = 1\), numerator is \(4(1)+8=12
eq0\). At \(x=-4\), numerator is \(4(-4)+8=-16 + 8=-8
eq0\). Wait, but the options have \(x = 1\) and \(x=-4\) as an option (the last option: \(x = 1\) and \(x=-4\)). Wait, but let's check the original problem again. Wait, maybe I made a mistake in factoring. Wait, the function is \(f(x)=\frac{4x + 8}{x^2+3x - 4}\). Let's re - factor the denominator: \(x^2+3x - 4\). Let's try to factor it as \((x + a)(x + b)\) where \(a + b = 3\) and \(ab=-4\). So \(a = 4\) and \(b=-1\), so \((x + 4)(x - 1)\). So denominator is zero when \(x=-4\) or \(x = 1\). Now, check numerator at \(x=-4\): \(4(-4)+8=-16 + 8=-8
eq0\). At \(x = 1\): \(4(1)+8=12
eq0\). So vertical asymptotes are at \(x = 1\) and \(x=-4\), which is the last option.

Wait, but let's check the options again:

Option 1: \(x=-1\) and \(x=-2\)

Option 2: \(x=-1\) and \(x = 2\)

Option 3: \(x=-1\) and \(x=-4\)

Option 4: \(x = 1\) and \(x=-4\)

So by factoring the denominator \(x^2+3x - 4=(x + 4)(x - 1)\), setting it to zero gives \(x=-4\) and \(x = 1\). And since the numerator is not zero at these points, these are the vertical asymptotes.

Step2: Verify numerator at critical points

For \(x = 1\): Numerator \(4x+8=4(1)+8 = 12
eq0\)

For \(x=-4\): Numerator \(4x + 8=4(-4)+8=-16 + 8=-8
eq0\)

So the vertical asymptotes are at \(x = 1\) and \(x=-4\)

Answer:

D. \(x = 1\) and \(x=-4\) (assuming the last option is D, but in the given options, the last option is "x = 1 and x = - 4")