Question

What is the zero of the function

Answer 1

Answer:

$x=6$ (or 6)

Step-by-step explanation:

Set f(x)=0:

$f(x)=\frac{x^2+x-42}{x^2+3x-28}$

$0=\frac{x^2+x-42}{x^2+3x-28}$

$0=x^2+x-42$

$0=(x+7)(x-6)$

$x=-7$ or $x=6$

Now check both solutions:

$f(-7)=\frac{(-7)^2+(-7)-42}{(-7)^2+3(-7)-28}$

$f(-7)=\frac{49-7-42}{49-21-28}$

$f(-7)=\frac{42-42}{28-28}$

$f(-7)=\frac{0}{0}$

Therefore, $x\neq-7$ because there is a hole at $(-7,0)$. The denominator, in addition, can never be 0, so the function is undefined for $x=-7$.

$f(6)=\frac{(6)^2+(6)-42}{(6)^2+3(6)-28}$

$f(6)=\frac{36+6-42}{36+18-28}$

$f(6)=\frac{42-42}{54-28}$

$f(6)=\frac{0}{24}$

$f(6)=0$

Since $f(6)=0$, then $x=6$ is the only zero of the function.