Question

find equations for the vertical asymptotes, if any, for the following rational function.
$f(x) = \frac{ -3x^2 + 13x + 10 }{ -3x + 7 }$
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Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(-3x^2 + 13x + 10\). We can factor it as \(-(3x^2 - 13x - 10)\). Using the quadratic formula or factoring, we get \(3x^2 - 13x - 10=(3x + 2)(x - 5)\), so the numerator is \(-(3x + 2)(x - 5)\). The denominator is \(-3x + 7\), which can be written as \(-(3x - 7)\).

Now, the function \(f(x)=\frac{-(3x + 2)(x - 5)}{-(3x - 7)}=\frac{(3x + 2)(x - 5)}{3x - 7}\) (after canceling the \(-1\) terms). Wait, no, actually, we should check for common factors first. Wait, the numerator is a quadratic and the denominator is linear. Let's check if the denominator's root makes the numerator zero. The root of the denominator \(-3x + 7 = 0\) is \(x=\frac{7}{3}\). Now, plug \(x = \frac{7}{3}\) into the numerator: \(-3(\frac{7}{3})^2+13(\frac{7}{3}) + 10=-3(\frac{49}{9})+\frac{91}{3}+10=-\frac{49}{3}+\frac{91}{3}+\frac{30}{3}=\frac{- 49 + 91+30}{3}=\frac{72}{3} = 24 eq0\). Wait, but wait, maybe I made a mistake in factoring? Wait, let's do polynomial long division. Divide \(-3x^2 + 13x + 10\) by \(-3x + 7\).

Dividing \(-3x^2+13x + 10\) by \(-3x + 7\):

\(-3x^2\div(-3x)=x\). Multiply \(-3x + 7\) by \(x\) to get \(-3x^2+7x\). Subtract from \(-3x^2 + 13x + 10\): \((-3x^2+13x + 10)-(-3x^2+7x)=6x + 10\). Now, \(6x\div(-3x)=-2\). Multiply \(-3x + 7\) by \(-2\) to get \(6x - 14\). Subtract: \((6x + 10)-(6x - 14)=24\). So, \(-3x^2 + 13x + 10=(-3x + 7)(x - 2)+24\). So, \(f(x)=x - 2+\frac{24}{-3x + 7}\).

Step2: Analyze vertical asymptotes

A vertical asymptote of a rational function \(y = \frac{N(x)}{D(x)}\) (where \(N(x)\) and \(D(x)\) are polynomials) occurs where \(D(x)=0\) and \(N(x) eq0\) at that point. But if we can simplify the function (i.e., there are common factors between numerator and denominator), then we have a hole instead of a vertical asymptote at the root of the common factor. But in our case, after trying to factor, we saw that the numerator is a quadratic and the denominator is linear, and they don't have a common factor (since the root of the denominator \(x=\frac{7}{3}\) does not make the numerator zero, as we calculated earlier). Wait, but wait, when we did polynomial long division, we saw that the function can be written as a linear function plus a rational function with non - zero remainder. Wait, no, actually, the key is: for a rational function, vertical asymptotes occur at the zeros of the denominator that are not zeros of the numerator. But first, we need to check if the numerator and denominator have any common factors. Let's check the degrees: numerator is degree 2, denominator is degree 1. So, they can't have a common factor (since a degree 1 polynomial and degree 2 polynomial can only have a common factor if the degree 1 polynomial divides the degree 2 polynomial. Let's check if \(-3x + 7\) divides \(-3x^2+13x + 10\). As we did polynomial long division, the remainder is 24, which is non - zero. So, the denominator \(-3x + 7 = 0\) when \(x=\frac{7}{3}\), and at \(x = \frac{7}{3}\), the numerator is \(-3(\frac{7}{3})^2+13(\frac{7}{3})+10=- \frac{49}{3}+\frac{91}{3}+10=\frac{42}{3}+10 = 14 + 10=24 eq0\). Wait, but wait, maybe I made a mistake in the initial thought. Wait, no, the formula for vertical asymptotes: if \(f(x)=\frac{N(x)}{D(x)}\), reduced form (no common factors), then vertical asymptotes at \(D(x)=0\). But in our case, when we simplify the function, is there a simplification? Wait, no, the numerator is quadratic, denominator is linear, no common factors. But wait, when we did the long division, we saw that \(f(x)=x - 2+\frac{24}{-3x + 7}\). The term \(\frac{24}{-3x + 7}\) has a vertical asymptote at \(x=\frac{7}{3}\)? Wait, no, wait, let's re - evaluate.

Wait, the denominator is \(-3x + 7\), set \(-3x+7 = 0\), we get \(x=\frac{7}{3}\). Now, we need to check if the numerator is zero at \(x=\frac{7}{3}\). Let's compute \(N(\frac{7}{3})=-3(\frac{7}{3})^2+13(\frac{7}{3})+10=-3\times\frac{49}{9}+\frac{91}{3}+10=-\frac{49}{3}+\frac{91}{3}+\frac{30}{3}=\frac{-49 + 91+30}{3}=\frac{72}{3}=24 eq0\). So, according to the definition, there should be a vertical asymptote at \(x = \frac{7}{3}\). Wait, but maybe I made a mistake in the long division. Wait, let's recalculate the long division:

Divide \(-3x^2+13x + 10\) by \(-3x + 7\):

The first term of the quotient: \(\frac{-3x^2}{-3x}=x\). Multiply \((-3x + 7)\) by \(x\): \(-3x^2+7x\). Subtract from \(-3x^2+13x + 10\): \((-3x^2+13x + 10)-(-3x^2+7x)=6x + 10\).

Second term of the quotient: \(\frac{6x}{-3x}=-2\). Multiply \((-3x + 7)\) by \(-2\): \(6x-14\). Subtract from \(6x + 10\): \((6x + 10)-(6x - 14)=24\). So, \(f(x)=x - 2+\frac{24}{-3x + 7}\). The function \(\frac{24}{-3x + 7}\) has a vertical asymptote at \(x=\frac{7}{3}\) because the denominator is zero at \(x = \frac{7}{3}\) and the numerator (24) is non - zero. So, the vertical asymptote is at \(x=\frac{7}{3}\). Wait, but the original problem's function: let's check again. The function is \(f(x)=\frac{-3x^2 + 13x + 10}{-3x + 7}\). So, denominator is \(-3x + 7\), set to zero: \(x=\frac{7}{3}\). Numerator at \(x=\frac{7}{3}\) is \(24 eq0\), so vertical asymptote at \(x=\frac{7}{3}\).

Wait, but maybe I made a mistake earlier. Let's confirm with the definition: A vertical asymptote is a vertical line \(x = a\) such that \(\lim_{x ightarrow a^+}f(x)=\pm\infty\) or \(\lim_{x ightarrow a^-}f(x)=\pm\infty\). For \(f(x)=\frac{-3x^2 + 13x + 10}{-3x + 7}\), as \(x ightarrow\frac{7}{3}^+\), the denominator \(-3x + 7 ightarrow0^-\) (since \(x>\frac{7}{3}\), \(-3x+7<0\)) and the numerator \(-3x^2 + 13x + 10 ightarrow - 3(\frac{7}{3})^2+13(\frac{7}{3})+10 = 24\) (a non - zero constant). So, \(\lim_{x ightarrow\frac{7}{3}^+}\frac{24}{-3x + 7}=-\infty\) (since numerator is positive and denominator is negative and approaching zero). Similarly, as \(x ightarrow\frac{7}{3}^-\), denominator \(-3x + 7 ightarrow0^+\) (since \(x<\frac{7}{3}\), \(-3x + 7>0\)), so \(\lim_{x ightarrow\frac{7}{3}^-}\frac{24}{-3x + 7}=+\infty\). So, there is a vertical asymptote at \(x=\frac{7}{3}\).

Wait, but the problem's interface has a "none" button. Did I make a mistake? Wait, let's re - factor the numerator. Let's use the quadratic formula for the numerator \(-3x^2+13x + 10\). The quadratic formula for \(ax^2+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a=-3\), \(b = 13\), \(c = 10\). So, \(x=\frac{-13\pm\sqrt{13^2-4\times(-3)\times10}}{2\times(-3)}=\frac{-13\pm\sqrt{169 + 120}}{-6}=\frac{-13\pm\sqrt{289}}{-6}=\frac{-13\pm17}{-6}\). So, \(x=\frac{-13 + 17}{-6}=\frac{4}{-6}=-\frac{2}{3}\) and \(x=\frac{-13 - 17}{-6}=\frac{-30}{-6}=5\). So, the numerator factors as \(-3(x+\frac{2}{3})(x - 5)=-(3x + 2)(x - 5)\) (multiplying \(-3\) and \((x+\frac{2}{3})\) gives \(-3x - 2\), wait, no: \(-3(x+\frac{2}{3})=-3x - 2\), so numerator is \((-3x - 2)(x - 5)=-(3x + 2)(x - 5)\). The denominator is \(-3x + 7=-(3x - 7)\). So, \(f(x)=\frac{-(3x + 2)(x - 5)}{-(3x - 7)}=\frac{(3x + 2)(x - 5)}{3x - 7}\). Now, the denominator is \(3x - 7\), set to zero: \(x=\frac{7}{3}\). The numerator at \(x=\frac{7}{3}\) is \((3\times\frac{7}{3}+2)(\frac{7}{3}-5)=(7 + 2)(\frac{7 - 15}{3})=9\times(-\frac{8}{3})=-24 eq0\). So, vertical asymptote at \(x=\frac{7}{3}\).

So, the equation of the vertical asymptote is \(x=\frac{7}{3}\).

Answer:

\(x=\frac{7}{3}\)