Question

The graph below shows the function f(x)=5x+10/x^2+7x+10 Where is the removable discontinuity of f(x) located?

Answer 1

The correct answer is x= - 2.

Step-by-step explanation:
A removable discontinuity happens where y=(0/0). To find this, we first factor the numerator and denominator.

For our numerator, we factor out the GCF; 5 is the GCF. Factoring 5 out of 5x leaves x, and factoring 5 out of 10 leaves 2, which gives us 5(x+2).

To factor the denominator, we want factors of c, 10, that sum to b, 7. 5*2=10 and 5+2=7, so this gives us (x+5)(x+2).

Next we find the common factors in the numerator and denominator; (x+2) is the common factor. We set this equal to 0: x+2=0. Now we solve for x by subtracting 2 from each side: x+2-2=0-2; x= - 2.

Answer 2

For this case we have the following function:

`f (x) = (5x + 10) / (x ^ 2 + 7x + 10)`
Let's rewrite the function:

`f (x) = (5x + 10) / ((x + 2) (x + 5)) f (x) = (5 (x + 2)) / ((x + 2) (x + 5))`
We set the denominator to zero to see the values of x for which it is not defined:

`(x + 2) (x + 5) = 0`
From here, we get:

`x = -2 x = -5`
There is a removable discontinuity at x = -2, since by rewriting the function we have:

`f (x) = 5 / (x + 5)`
Answer:
the removable discontinuity of f (x) is located at:
x = -2