Question

Help please i will give you medals!!! I need help asap please!!!!

What is the value is a discontinuity of x^2+2x+3 over x^2-x-12

x=-1
x=-2
x=-3
x=-4

Answer 1

Factor x^2+2x+3 to get (x+1)(x+2)


Factor x^2-x-12 to get (x-4)(x+3)


Nothing simplifies/cancels, so we focus on the denominator now.


The denominator is factored to be (x-4)(x+3). Set it equal to zero and solve for x


(x-4)(x+3) = 0

x-4=0 or x+3=0

x=4 or x=-3

Those last two values, x=4 or x=-3, make the denominator zero...but remember: you CANNOT divide by zero.


So those values cause a discontinuity

Answer 2

@whpalmer4

Answer 3

so is it -3?

Answer 4

correct

Answer 5

can you help me with another one?

Answer 6

sure

Answer 7

Find the horizontal or oblique asymptote of f(x) = negative 3 x squared plus 7 x plus 1, all over x minus 2

y = 2

y = -3

y = 3x + 7

y = -3x + 1

Answer 8

Are you familiar with polynomial long division?

Answer 9

yes

Answer 10

for a rational, a "discontinuity" will be at the point where its asymptotes are

as pointed out by jim_thompson5910 , you can get the horizontal asymptotes by getting the "zeros" of the denominator, that is
\(\bf x^2-x-12 = 0\)

at those points, the denominator will be zero, thus the fraction "undefined", thus will yield a "discontinuity"

Answer 11

@jim_thompson5910 and @jdoe0001 have really said it all...

Answer 12

please note there was an error in the first line
x^2 + 2x + 3 is not (x + 1)(x+2)
it did not affect the denominator