Question 1137488
here's your graph.


<img src = "http://theo.x10hosting.com/2019/032701.jpg" alt="$$$" >


the equation that was graphed is y = x^2/(x^2+x-12)


you have vertical asymptotes at x = -4 and x = 3.


that's because x^2 + 2x - 12 factors into (x+4) * (x-3.


the denominator becomes 0 when x = -4 and x = 3.


those value of x are where your vertical asymptotes are.


since no real value of y is possible when x = -4 and x = 3, those values of x are not in the domain.


therefore, the domain is all real values of x except at x except at x = -4 and x = 3.


the range is all real values of y.


the x intercept is the value of x when the value of y is equal to 0.


to find those values of x, set y = 0 and solve for x.


y = x^3 / (x^2 + x - 12) becomes x^3 / (x^2 + x - 12) = 0 when y = 0.


multiply both sides of the equation by (x^2 + x - 12) and you get x^3 = 0


solve for x to get x = 0.


that's your x-intercept.


to solve for the y-intercept, make x = 0 and you get y = x^3 / (x^2 + x - 12) becomes y = 0 / (-12) which becomes y = 0.


the y-intercept is at x = 0.


your x-intercept is at (0,0).


your y-intercept is at (0,0).