Respuesta :
Answer:
The zeros of our function f is at x=-1.
The discontinuity is at x=-4.
These are correct if the function is [tex]f(x)=\frac{x^2+5x+4}{x+4}[/tex] .
Please let know if I did not interpret your function correctly.
Step-by-step explanation:
I imagine you mean [tex]f(x)=\frac{x^2+5x+4}{x+4}[/tex] but please correct me if I'm wrong.
The zero's of a rational expression occur from it's numerator.
That is, in a fraction, the only thing that makes that fraction 0 is it's numerator.
So we need to solve [tex]x^2+5x+4=0[/tex] for x.
The cool thing is this one is not bad to factor since the coefficient of x^2 is 1. When the coefficient of x^2 is 1 and you have a quadratic, all you have to do is ask yourself what multiplies to be c and adds to be b.
[tex]x^2+5x+4[/tex] comparing to [tex]ax^2+bx+c[/tex] gives you [tex]a=1,b=5,c=4[/tex].
So we are looking for two numbers that multiply to be c and add to be b.
We are looking for two numbers that multiply to be 4 and add to be 5.
Those numbers are 1 and 4 since 1(4)=4 and 1+4=5.
The factored form of [tex]x^2+5x+4[/tex] is [tex](x+1)(x+4)[/tex].
So [tex]x^2+5x+4=0[/tex] becomes [tex](x+1)(x+4)=0[/tex].
If you have a product equals 0 then at least one of the factors is 0.
So we need to solve x+1=0 and x+4=0.
x+1=0 when x=-1 (subtracted 1 on both sides to get this).
x+4=0 when x=-4 (subtracted 4 on both sides to get this).
The zeros of our function f is at x=-1 and x=-4.
Now to find where it is discontinuous. We have to think 'oh this is a fraction and I can't divide by 0 but when is my denominator 0'. If the value for the variable is not obvious to you when the denominator is 0, just solve x+4=0.
x+4=0 when x=-4 (subtracted 4 on both sides).
So we have a contradiction at one of the zeros so x=-4 can't be a zero.
The discontinuity is at x=-4.
Answer:
This function is discontinuous at x = 4, and has a zero at x = -1.
Step-by-step explanation:
If x = -4, the denominator will be zero and thus the function will be undefined. Thus, the discontinuity is at x = -4.
To find the zero(s): Set the numerator = to 0, obtaining
x^2+5x+4 = 0. Factoring this, we get (x + 4)(x + 1) = 0. Thus, we have a zero at x = -1.
Notice that f(x) can be rewritten as
x^2 + 5x + 4 (x+4)(x+1)
f(x) = -------------------- = ---------------- = x + 1 for all x other than x = -4.
x + 4 (x+4)
This function is discontinuous at x = 4, and has a zero at x = -1.
