Asked by sh
At what points on the curve y=(x^2)/(2x+5) is the tangent line horizontal?
y'=[(2x)(2x+5)-(2)(x^2)]/(2x+5)^2
=[2x^2+10x]/(2x+5)^2
I plugged 0 into x, and got the coordinates (0,0). How do I get the other pair of coordinates?
Thanks in advance.
y'=[(2x)(2x+5)-(2)(x^2)]/(2x+5)^2
=[2x^2+10x]/(2x+5)^2
I plugged 0 into x, and got the coordinates (0,0). How do I get the other pair of coordinates?
Thanks in advance.
Answers
Answered by
jim
Just from looking at 2x^2+10x and that has to be zero - I'd try -5
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