grangergirl
Answered

Create an equation!


A graph has a y-intercept at -5, no x-intercepts, and discontinuous points at (-1,-5) and (3, -5). I want to form an equation for this graph, but I don't know how the y-intercept relates to the graph of a rational function. Here's all I have so far:

y = (x+1)(x-3)/(x+1)(x-3)

I realize this isn't even close to being complete; I need some direction on how to finish the equation. Any help would be appreciated. Thanks.



Answer :

paulcox
that is close: you have the intercepts correct and the discontinuous point at x=-1 and x=3. The only thing is that you need to fix is the y values of the discontinuous points.

Without any coefficients, your graph is basically y=1, and you need it to resemble y=-5 to get the right y values. Normally, that just involves multiplication by -5 so multiply your function by -5 and you should get a correct graph.

And make sure to denote it with y= since it is an equation