which is correct?

[allegansveritatem]

I came across this paradox (?) while working out a graph of a rational function. I was looking for the y intercept. I worked it out thus:


I looked up the solution and found that I had gotten it wrong and this is the way the solutions manual had it worked out:



I suspect this discrepancy has something to do with f(0) being a function....but what's happening here?

 

[Dr.Peterson]

What you calculated the first time was not the y-intercept, but the x-intercept. Did you notice that what you got there was a value of x, not of y?

Both are important, but they are very different in their roles.

An x-intercept is a point (there may be many) where the graph crosses the x-axis: that is, the value of x where y=0. You set f(x) = 0 and solve. That's what you did the first time.

The y-intercept is the point (there can be only one for a function) where the graph crosses the y-axis. You set x = 0 and find y: that is, you evaluate f(0).

No paradox; you just answered the wrong question.

 

[pka]

I came across this paradox (?) while working out a graph of a rational function. I was looking for the y intercept. I worked it out thus:
View attachment 13869
I looked up the solution and found that I had gotten it wrong and this is the way the solutions manual had it worked out:
I suspect this discrepancy has something to do with f(0) being a function....but what's happening here?

Look at this webpage. The very first thing to do is to plot the graph.
That web site is free to use (if you like you can go pro)
Even without any of that, just by looking at the problem you can see that \(\displaystyle x-1\) gives a \(\displaystyle x-\)intercept.
We see \(\displaystyle x=-2~\&~x=3\) give vertical asymptote.

 

[allegansveritatem]

What you calculated the first time was not the y-intercept, but the x-intercept. Did you notice that what you got there was a value of x, not of y?

Both are important, but they are very different in their roles.

An x-intercept is a point (there may be many) where the graph crosses the x-axis: that is, the value of x where y=0. You set f(x) = 0 and solve. That's what you did the first time.

The y-intercept is the point (there can be only one for a function) where the graph crosses the y-axis. You set x = 0 and find y: that is, you evaluate f(0).

No paradox; you just answered the wrong question.

well, what I must have been thinking is this....well when it comes down to saying what I was thinking about I can't come up with anything. All I can say is: It seemed like the way to go at the time. But now I see that I very obviously got something that x is equal to, not y. f(0) means: What do you get when you set x to zero. And what else but the point where x is directly above or below the place where the graph intersects with the y axis. Oh boy! Next time I'll wait til I wake up before I do my math work. Thanks

 

[allegansveritatem]

Look at this webpage. The very first thing to do is to plot the graph.
That web site is free to use (if you like you can go pro)
Even without any of that, just by looking at the problem you can see that \(\displaystyle x-1\) gives a \(\displaystyle x-\)intercept.
We see \(\displaystyle x=-2~\&~x=3\) give vertical asymptote.

that Wolfram Alpha is a good site. As for going pro...I am of the old school that likes to buy things ONE TIME. That is not the business model used anymore it seems..but do you remember when you bought a radio and that was that? Or you bought a computer program and that was that? Not anymore thank you Steve Jobs. Oh well, enough of that.