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# find slopes of all lines tangent to the graph y^3 + 5x^15y^2 - 50y = 150 - 150x^5 at points on its graph were x = 1?

one of the slope is m = 15

there are two more.

idk how to solve this problem. i try to take the derivative but i dont really know what its asking from me. can someone solve it so that i can see what happene?

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- alexLv 71 month ago
x=1 --->y^3+5y^2-50y=0---> y = 0 , -10 , 5

3 points (1,0) , (1,-10) and (1,5)

implicit differentiate f(x,y)

1 of the point gives the slope 15

(you can continue)

- DixonLv 71 month ago
The question is basically asking what is the slope of the curve at x = 1. A line tangent to the curve is at the slope of the curve where it touches.

The more tricky part looks to be getting dy/dx at all and it is a very strange shape curve that looks to have two or three places where x = 1.

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