Write rational function equation given graph

[Coco12]

On The graph, the following information was given:
Point of discontinuity( -2,0) and (1,-3)
X intercept at ( 2,0)
Yintercept (0,-4)

This is what I had : (x+2)(x-2)(x-1)/ (x+2)(x-1)

But the answer is that u have to put (x+2)^2 in the numerator instead of what I had.
But I thought that the point of discontinuity was on that spot ( xintercept) and thus it would not be counted that there was an xintercept AND a point of discontinuity at that spot. Can someone explain this to me?

 

[pbuk]

What is the y-intercept of your graph?

 

[LCKurtz]

On The graph, the following information was given:
Point of discontinuity( -2,0) and (1,-3)
X intercept at ( 2,0)
Yintercept (0,-4)

This is what I had : (x+2)(x-2)(x-1)/ (x+2)(x-1)

But the answer is that u have to put (x+2)^2 in the numerator instead of what I had.
But I thought that the point of discontinuity was on that spot ( xintercept) and thus it would not be counted that there was an xintercept AND a point of discontinuity at that spot. Can someone explain this to me?

Hopefully you were asked to find a solution, not the solution, because there is more than one answer. Also the question doesn't specifically rule out vertical asymptotes for the discontinuities. Even if they are ruled out, there is still more than one answer. But to answer your question, it wouldn't matter whether there was $\frac{(x+2)^2}{x+2}$ or $\frac{x+2}{x+2}$. They both have a removable discontinuity at $(-2,0)$. That point is not on the graph in either case, so it is not an intercept.

[Edit, added]: Actually the seond one's removable discontinuity is at $(-2,1)$ which is likely why your teacher wanted the numerator factor squared.

 

[haruspex]

Point of discontinuity( -2,0) and (1,-3)

I don't understand the significance of quoting a y value for a point of discontinuity in a rational function of a single variable. Is there any more information? An actual graph perhaps?

 

[LCKurtz]

I don't understand the significance of quoting a y value for a point of discontinuity in a function of a single variable. Is there any more information? An actual graph perhaps?

I'm guessing they want a removable discontinuity at that point.

 

[Coco12]

Is it because, you need for it to be squared in order for the y intercept to be -4?

 

[LCKurtz]

Is it because, you need for it to be squared in order for the y intercept to be -4?

Please use the quote button so we know who you are asking, and what "it" is. If you are referring to the $(y+2)^2$, the answer is no. It is to make (-2,0) be where the hole on the graph is.

 

[Coco12]

Please use the quote button so we know who you are asking, and what "it" is. If you are referring to the $(y+2)^2$, the answer is no. It is to make (-2,0) be where the hole on the graph is.

i mean for the numerator, because the way to find y intercept is to make all the x equal to zero

 

[LCKurtz]

Here's a picture of your graph after you square the $x+2$ giving$$
f(x) = \frac{(x+2)^2(x-2)(x-1)}{(x+2)(x-1)}$$That graph is the green parabola with the holes at $(-2,0)$ and $(1,-3)$. The red graph is$$
g(x) = \frac{-4(x-2)}{(x-1)(x+2)}$$It has the same $x=2$ and $y=-4$ intercepts and is also discontinuous at $x=-2$ and $x = 1$ by virtue of having vertical asymptotes there.

 

[Coco12]

Here's a picture of your graph after you square the $x+2$ giving$$
f(x) = \frac{(x+2)^2(x-2)(x-1)}{(x+2)(x-1)}$$That graph is the green parabola with the holes at $(-2,0)$ and $(1,-3)$. The red graph is$$
g(x) = \frac{-4(x-2)}{(x-1)(x+2)}$$It has the same $x=2$ and $y=-4$ intercepts and is also discontinuous at $x=-2$ and $x = 1$ by virtue of having vertical asymptotes there.

The graph looks like the green one. So I guess the reason why it would be squared is only for the y intercept

 

[Mentallic]

The graph looks like the green one. So I guess the reason why it would be squared is only for the y intercept

No, it's to have an x-intercept at x=-2.

\frac{(x+2)(x-2)(x-1)}{(x+2)(x-1)}

has a removable discontinuity at x=-2 (and x=1 but let's ignore this for a sec). If we cancel the common factors from the numerator and denominator then we're left with x-2 and plugging in x=-2 gives us -4. But we want (-2,0) to be a point of discontinuity, and not (-2,-4). The only way we can do this is if we end up with (x+2)p(x) where p(x) is any polynomial after cancelling all common factors, and the only way we can achieve this is by squaring the (x+2) factor.

All the information you're given was chosen to fit on a parabola (with discontinuities of course). For this reason, you don't even need to multiply your final expression by some constant.

So what I mean by this and how you would go about solving this problem is that since you're told an x-intercept is at x=2, but also that an x-intercept is at x=-2 since (-2,0) satisfies the equation after being simplified, from this information alone you know it needs to be of the form (x+2)(x-2) after simplification, and since it's a point discontinuity at x=-2, you also need to include the factors (x+2)/(x+2). Now you have

\frac{(x+2)^2(x-2)}{x+2}

But you're also told that a point discontinuity occurs at x=1, so the factors (x-1)/(x-1) also need to be included. Hence we have

\frac{(x-1)(x+2)^2(x-2)}{(x-1)(x+2)}

But we didn't even consider that the y-value of the point discontinuity at x=1 is -3. Well, is it? We plug in x=1 AFTER simplifying (because with the (x-1) factors we'd end up with 0/0) and we can also cancel out an x+2 factor from the numerator and denominator since we'd end up multiplying and dividing by the same number. So we have

(x+2)(x-2)

and plugging in x=1 gives -3

So it turns out that (1,-3) happens to lie on the parabola, hence we don't need a constant factor. For argument's sake, let's consider if this weren't the case. Say, if when we plugged in x=1 we ended up with y=-6, then since we need it to be y=-3 then we'd divide the rational function by 2 to get

\frac{(x-1)(x+2)^2(x-2)}{2(x-1)(x+2)}


Also finally, if we plug x=0, we're given the y-intercept and it also turns out to be y=-4, so the question was clearly constructed to work nicely given that you construct the rational function from the x-intercepts and point discontinuities only.

 

[pbuk]

The graph looks like the green one. So I guess the reason why it would be squared is only for the y intercept

Yes indeed - there appears to be some selective deafness here

 

[LCKurtz]

The graph looks like the green one. So I guess the reason why it would be squared is only for the y intercept

Yes indeed...

The main reason the x+2 is squared is to get the "intercept" at x=-2. Except that it isn't an intercept; it is a removable discontinuity. The fact that the y intercept is -4 just happens.

Personally, I think the teacher started with $f(x) = x^2-4$ and then threw in some factors to make the two holes in the graph. But the conditions $f(0)=-4$ and $f(1)=-3$ are redundant. You wouldn't normally expect a parabola to pass through two specified intercepts plus two more specified points.

Also my point in showing the two graphs is to show the solution isn't unique or, at the minimum, the problem is poorly thought out.