Question

what is the oblique asymptote of the funtion f(x)=(x^2-5x+6)/(x+4)

Answer 1

I got y=x-1 is that right?

Answer 2

no

Answer 3

dividing (x^2-5x+6)/(x+4) by (x+4) yields x - 9 + 36/(x + 4), no?

So your oblique asymptote is y = x - 9, no?

Answer 4

Your right thank you. =D

Answer 5

can you help me with this one too, what are the zeros of the function
f(x)=3x^2-9x+6/3x-3

Answer 6

I got 1

Answer 7

You do realize that a zero of the polynomial is the same thing as an x-intercept. You are being asked to find all values of x where the function crosses the x-axis, that is where the value of the function is zero.

So you want to solve for x, when f(x) = 0.

f(x)=3x^2-9x+6/3x-3 = 0

In other words, solve the equation 3x^2 - 9x + 6/(3x - 3) = 0.

Can you do that?

Answer 8

-1

Answer 9

or is it 2 or -2........IDK

Answer 10

The zeros of the function occur when y is 0. So you can write:
\[\frac{3x^2-9x+6}{3x-3}=0\]

Answer 11

However you then realize that a fraction is 0 ONLY IF its numerator is 0 so you write:
\[3x^2-9x+6=0\]

Answer 12

Solve that by factoring out the 3:
\[3(x^2-3x+2)=0\]

Answer 13

And then factor the trinomial:
\[3(x-2)(x-1)=0\]

Answer 14

Set each factor equal to 0 and solve. Obviously 3 cannot be 0 so x=2 and x = 1 are zeros of the function.

Answer 15

It looked like your equation was f(x)=3x^2-9x+6/(3x-3).

I guess you meant f(x)=(3x^2-9x+6)/(3x-3).

In that case, Mertsj has given the correct answer.