SOLUTION: A curve has the equation y=(x-3)^2/(x+1). By considering y for all real values of X, show that no point on the curve exists in the interval -16<y<0.

Algebra ->  Graphs -> SOLUTION: A curve has the equation y=(x-3)^2/(x+1). By considering y for all real values of X, show that no point on the curve exists in the interval -16<y<0.      Log On


   

Question 1059841: A curve has the equation y=(x-3)^2/(x+1). By considering y for all real values of X, show that no point on the curve exists in the interval -16
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
-16 is not an interval. The best that can be done is look at the actual intervals, according to critical x values of your equation. Those are at 3 and -1. Obviously no point happens for x=-1 because your equation is undefined there (and would make for division by zero).

Now the intervals to examine are -infinity%3Cx%3C-1 and -1%3Cx%3C=3 and 3%3C=x%3Cinfinity. You can pick ANY x-value in these three intervals and there will be a corresponding y value. The x=-16 WILL give a real value for y, and is in the first interval listed.

NOTE: Intervals are generally for parts of the x-axis. We do not ordinarily refer to intervals on the y-axis, although we might if we want. A quick view of your equation's graph:



graph%28400%2C400%2C-8%2C8%2C-16%2C18%2C%28x-3%29%5E2%2F%28x%2B1%29%29
Or better,
graph%28400%2C400%2C-2%2C14%2C-2%2C32%2C%28x-3%29%5E2%2F%28x%2B1%29%29
There is a vertical asymptote at x=-1. Check the sign algebraically for the interval -infinity%3Cx%3C-1. Whatever you find, your graph will have at least the local minimum at what looks like (3,0), and the value of y in at least the middle interval on x certainly gives y values greater than 16.

Another graph look:
graph%28400%2C400%2C-20%2C3%2C-20%2C8%2C%28x-3%29%5E2%2F%28x%2B1%29%29

Just using the graphing feature or any good graphing tool, it seems there is no value for y in -16%3C=y%3C=0. I have not examined this much using algebra.