Grangergirlidn
Answered

Discover new perspectives and gain insights with IDNLearn.com's diverse answers. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.

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.

Sagot :

Paulcoxidn
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
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.