Find The equation of the tangent to the curve y=x^2+x^3 at the point (1,0) .

Question

ash2326 · Answer

@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
$$\frac{(y-y1)}{(x-x1)}=m$$
where 
$$(x1,y1)=(1, 0)
$$
and
$$\large m= ( \frac{dy}{dx})_{(1,0)}$$
Can you try now?

anonymous · Answer

so its gonna be $$\frac{y-0}{x-1}=0$$ ?

anonymous · Answer

@ash2326 .

ash2326 · Answer

$$\frac{dy}{dx}=3x^2+2x$$
put x=1 and y=0
$$\frac{dy}{dx}=3+2=5$$

anonymous · Answer

Oh Sry i think i need some sleep xD, U did great :D
Although its $$2x-3x^2  ...$$ where the slope is -1 
TY ^_^

ash2326 · Answer

You really are sleepy:P
you wrote
y=x^2+x^3 instead of y=x^2-x^3

anonymous · Answer

OH LOL ,Its  y=x^2-x^3 .. Night :D