SOLUTION: Given p(x) = x^3+5x^2+4x/3(x+1)(x^2−7x+12), find: (a) X-intercept(s) (b) Y-intercept (c) Vertical asymptote(s) (d) Hole(s) (e) End behavior / horizontal asymptote

Question 1167431: Given p(x) = x^3+5x^2+4x/3(x+1)(x^2−7x+12),
find:
(a) X-intercept(s)
(b) Y-intercept
(c) Vertical asymptote(s)
(d) Hole(s)
(e) End behavior / horizontal asymptote
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Your function "x^3+5x^2+4x/3(x+1)(x^2−7x+12)" means this:

If you are working on a problem like this, you should know that parentheses in the right places are important....

Undoubtedly the function is supposed to be "(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12))"

Now we can answer all the questions quickly by factoring numerator and denominator:

(a) x-intercept(s): wherever there is a linear factor in the numerator without a like factor in the denominator. (x=0, x=-4)

(b) y-intercept: (0,b) where b = p(0). Since x is a factor of the numerator, the y-intercept is (0,0).

(c) Vertical asymptote(s): wherever there is a linear factor in the denominator without a like factor in the numerator. (x=3, x=4)

(d) Hole(s): wherever there are like linear factors in both numerator and denominator. (x=-1)

(e) End behavior/horizontal asymptote: The degrees of numerator and denominator are the same, so the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator. (y = 1/3)

A graph showing the x- and y-intercepts....

Another showing the horizontal asymptote y=1/3....

And one showing the effect of the vertical asymptotes at x=3 and x=4....

And finally one showing the hole at (-1,-0.5)....

The hole will show up better on a graphing utility like a TI83 calculator...

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