Question

what is te discontinuity of the function f(x) = the quantity of x squared plus 6x plus 8 all over x plus 4?

Answer 1

Answer:

A hole at x=-4.

Step-by-step explanation:

This is a fraction so we have to worry about division by zero.

The only time we will be dividing by 0 is when x+4 is 0.

Solving the equation

x+4=0 for x:

Subtract 4 on both sides:

x=-4

So there is either a vertical asymptote or a hole at x=-4.

These are the kinds of discontinuities we can have for a rational function.

If there is a hole at x=-4, then x=-4 will make the top zero and can be cancelled out after simplification.

If is is a vertical asymptote, x=-4 will make the top NOT zero.

Let's see what -4 for x in x^2+6x+8 gives us:

(-4)^2+6(-4)+8

16+-24+8

-8+8

0

Top and bottom are 0 when x=-4.

Let's see what happens after simplication.

We are going to factor a^2+bx+c if factorable by finding two numbers that multiply to be c and add up to be b.

So what 2 numbers together multiply to be 8 and add up to be 6.

I hoped you said 4 and 2 because (4)(2)=8 where 4+2=6.

$\frac{x^2+6x+8}{x+4}=\frac{(x+4)(x+2)}{x+4}=x+2$

We we able to cancel out that factor that was giving us x=-4 is a zero.  

Therefore there is a hole at x=-4.