the graph below shows the function $f(x)=\frac{5x + 10}{x^{2}+7x + 10}$. where is the removable…

the graph below shows the function $f(x)=\frac{5x + 10}{x^{2}+7x + 10}$. where is the removable discontinuity of $f(x)$ located? $x=-5$

the graph below shows the function $f(x)=\frac{5x + 10}{x^{2}+7x + 10}$. where is the removable discontinuity of $f(x)$ located? $x=-5$

Answer

Explanation:

Step1: Factor the function

First, factor the numerator and denominator. The numerator $5x + 10=5(x + 2)$. The denominator $x^{2}+7x + 10=(x + 2)(x+5)$. So, $f(x)=\frac{5(x + 2)}{(x + 2)(x + 5)}$.

Step2: Identify the removable - discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x + 2)$ cancels out (for $x\neq - 2$). The value of $x$ that makes the cancelled - out factor equal to zero is the location of the removable discontinuity. Setting $x+2 = 0$, we get $x=-2$.

Answer:

$x=-2$