James, I gave you a detailed solution yesterday for the same question on the assumption that your symbol was supposed to be ≥
You later told me that it was to be ≤
So why don't you just switch all the conditions
e.g. all the domain that worked before, now will not
and the domain that did not work, now will
you could also follow my Wolfram link and change the ≥ to ≤ , which will reflect the graph in the x-axis.
I am sure you can do that.
http://www.jiskha.com/display.cgi?id=1396018412
Solve the following rational inequality. Write the solution in interval notation. Show the number line and breaking points or graph of the polynomial with the part of the graph that is the solution (x-4)(x+2)/(x-3)(x+5) ¡Ü 0
2 answers
I can't make out the inequality sign, but that really makes no difference to the solution
y = (x-4)(x+2)/(x-3)(x+5)
You know there are roots at -2 and 4, and asymptotes at -5 and 3.
So, the graph will cross the x-axis at -2 and 4.
When x < -5, all factors are negative, so y > 0 : --/-- = +
for -5 < x < -2, --/-+ = -
for -2 < x < 3, -+/-+ = +
for 3 < x < 4, -+/++ = -
for x > 4, ++/++ = +
To verify this, check the graph at
http://www.wolframalpha.com/input/?i=%28%28x-4%29%28x%2B2%29%29%2F%28%28x-3%29%28x%2B5%29%29+for+x+%3D+-7..5
y = (x-4)(x+2)/(x-3)(x+5)
You know there are roots at -2 and 4, and asymptotes at -5 and 3.
So, the graph will cross the x-axis at -2 and 4.
When x < -5, all factors are negative, so y > 0 : --/-- = +
for -5 < x < -2, --/-+ = -
for -2 < x < 3, -+/-+ = +
for 3 < x < 4, -+/++ = -
for x > 4, ++/++ = +
To verify this, check the graph at
http://www.wolframalpha.com/input/?i=%28%28x-4%29%28x%2B2%29%29%2F%28%28x-3%29%28x%2B5%29%29+for+x+%3D+-7..5