Question 1167431
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Your function "x^3+5x^2+4x/3(x+1)(x^2−7x+12)" means this:<br>
{{{x^3+5x^2+4x/3(x+1)(x^2-7x+12)}}}<br>
If you are working on a problem like this, you should know that parentheses in the right places are important....<br>
Undoubtedly the function is supposed to be "(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12))"<br>
{{{(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12))}}}<br>
Now we can answer all the questions quickly by factoring numerator and denominator:<br>
{{{((x)(x+1)(x+4))/((3)(x+1)(x-3)(x-4))}}}<br>
(a) x-intercept(s): wherever there is a linear factor in the numerator without a like factor in the denominator.  (x=0, x=-4)<br>
(b) y-intercept: (0,b) where b = p(0).  Since x is a factor of the numerator, the y-intercept is (0,0).<br>
(c) Vertical asymptote(s): wherever there is a linear factor in the denominator without a like factor in the numerator.  (x=3, x=4)<br>
(d) Hole(s): wherever there are like linear factors in both numerator and denominator.  (x=-1)<br>
(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)<br>
A graph showing the x- and y-intercepts....<br>
{{{graph(400,400,-10,10,-1,2,(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12)))}}}<br>
Another showing the horizontal asymptote y=1/3....<br>
{{{graph(400,400,-100,100,-1,2,(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12)))}}}<br>
And one showing the effect of the vertical asymptotes at x=3 and x=4....<br>
{{{graph(400,400,-10,10,-50,50,(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12)))}}}<br>
And finally one showing the hole at (-1,-0.5)....<br>
{{{graph(400,400,-1.001,-0.999,-.1,.1,(x^3+5x^2+4x)/(3(x+1)(x^2-7x+12)))}}}<br>
The hole will show up better on a graphing utility like a TI83 calculator...<br>