# A function is a ratio of quadratic functions and has a vertical asymptote x =4 and just one -intercept, x =1. It is known f that has a removable discontinuity at x =- 1 and lim_(x->1)f(x)=2. Evaluate a) f(0) b) limx_(x->oo)f(x)

A function is a ratio of quadratic functions and has a vertical asymptote x =4 and just one -intercept, x =1. It is known f that has a removable discontinuity at x =- 1 and $\underset{x\to 1}{lim}f\left(x\right)=2$. Evaluate
a) f(0)
b) $lim{x}_{x\to \mathrm{\infty }}f\left(x\right)$
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Nathaniel Kramer

When it says that it is a ratio of quadratics that means that the numerator and the denominator are polynomials of the second degree. This means that they both have two roots. To find vertical asymptotes you set the denominator equal to zero and solve for x. Stating the vertical asymptotes gives you a root of the polynomial in the denominator. The part about it having one x intercept gives you a root of the polynomial in the numerator. The part about the removable discontinuity means that the two polynomials share a common root and the limit at that discontinuity gives us an idea about how to find a constant. In our case the top polynomial (without the constant that we will determine in a little bit) is $\left(x-1\right)\left(x+1\right)$. the bottom polynomial is $\left(x-4\right)\left(x+1\right)$. where $x=-1$ is the common root, $x=4$ is the vertical asymptote and $x=1$ is the x intercept. So, we have
$f\left(x\right)=\frac{c\left(x-1\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}$ where c is some constant
$\underset{x\to -1}{lim}\frac{c\left(x-1\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}=\underset{x\to -1}{lim}\frac{c\left(x-1\right)}{x-4}$
$=\frac{c\left(1-1\right)}{-1-4}$
$=\frac{-2c}{-5}$
$=2$
So, $c=5\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{5\left(x-1\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}$
a) $f\left(0\right)=\frac{5}{4}$
b) The degree of the two polynomials is the same so the limit is 5.