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ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1@Eyad We have to find the equation of the tangent to the curve \(y=x^2+x^3\) at the point (1, 0) First find \(\frac{dy}{dx}\) at the point (1, 0) this will be the slope of the tangent at (1, 0) and then you can find the equation using this \[\frac{(yy1)}{(xx1)}=m\] where \[(x1,y1)=(1, 0) \] and \[\large m= ( \frac{dy}{dx})_{(1,0)}\] Can you try now?

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0so its gonna be \[\frac{y0}{x1}=0\] ?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dx}=3x^2+2x\] put x=1 and y=0 \[\frac{dy}{dx}=3+2=5\]

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0Oh Sry i think i need some sleep xD, U did great :D Although its \[2x3x^2 ...\] where the slope is 1 TY ^_^

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1You really are sleepy:P you wrote y=x^2+x^3 instead of y=x^2x^3

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0OH LOL ,Its y=x^2x^3 .. Night :D
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