Posted by Amy on Friday, December 9, 2011 at 4:51pm.
What woman? It's just a math problem, like others in your book...
You have (x-4)(x+3)/(x+1)
domain is all reals except where denominator is zero. That is all reals except x = -1
zeros at x = 4 and -3 because that's where the numerator is zero and the denominator is not zero
no horizontal asymptotes, since the numerator has higher degree than the denominator. As x gets large, the fraction is just x^2/x = x
So, oblique asymptote is y=x
The graph intersects it at (-6,-6)
A little interval chart will show that
y<0 for x in (-oo,-3)U(-1,4)
y>0 for x in (-3,-1)U(4,oo)
At the end you say you can graph it. Well! In that case, what's the problem? The graph shows all the answers to the questions.
Related Questions
Pre-Cal - I'm trying to put this parabola in standard form which is: (y-k)^2...
pre cal - convert the point to polar coordinates: (12,-12)
pre cal - does anyone take pre-cal online ?
pre - cal - 12-5y^2 what's the question?
Pre-cal - Does the equation x^2+x-12/x+2 have holes?
Pre-Cal. - f(x)=x^3-3x^2+x-12 =x^2(x-3)+4(x-3) =x^2+4) (x-3) so if f(x) = 0 at x...
Pre-cal - let P(x)=2x^4-x^3+13^2-8x-24. Use synethetic division to show that x+1...
Pre-Cal - Given that lim x-> f(x)=6 and that lim x-> g(x)= -4, ...
Pre-Cal. - find all the zeros of f(x)=x3-3x2+4x-12
Pre cal - Find the area of a triangle with sides of 12, 14 and 20
For Further Reading