Question

Calculus question:

graph y= (x^2 +2x + 16)/(X+2)

rewritten form:

y'

y' in single fraction:

critical numbers

critical points

graph:

Please help me in my homework. Thanks!

Answer 1

\[\frac{ x^2 +2x + 16 }{ x + 2 }\]

^rewritten equation

Answer 2

not really sure what calculus you need for this
you have a vertical asymptote at the zero of the denominator, namely at \(x=-2\)
there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator, but if you do the division you get
\[f(x)=x+\frac{16}{x+2}\] so you also have a slant asymptote at \(y=x\)

Answer 3

did you find \(y'\) ? it requires the quotient rule

Answer 4

Not yet. Can you show me? Just so I'm sure what's right. Thanks!

Answer 5

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=x^2+2x+16, f'(x)=2x+2, g(x)=x+2, g'(x)=1\]

Answer 6

\[f'(x) = \frac{x^2+4x-12}{(x+2)^2}\]
now, I break it down into 2 parts: numerator and denominator
1/ numerator: A = \(x^2 +4x-12=(x-2)(x+6) \) therefore, at x = 2 and x = -6 A =0
I have a table for A (numerator)
it means out of the interval of [-6,2] numerator A >0 , and among [-6,2] , numerator A <0
Got this part?

Answer 7

2/ for denominator, let B = denominator = (x+2) ^2 . At x = -2 B =0 which makes f'(x) undefined. Except this points, no matter what value of x is, B >0 , right? because
B = (x+2 )^2
so, I have a table for B, like
got this part?

Answer 8

Now, I combine A /B . That is f'(x) to consider how the graph look like
so, I have the sign of the f'(x) and critical points. Now consider f(x)