## anonymous 4 years ago Find The equation of the tangent to the curve y=x^2+x^3 at the point (1,0) .

1. ash2326

@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?

2. anonymous

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

3. anonymous

@ash2326 .

4. ash2326

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

5. anonymous

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

6. ash2326

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

7. anonymous

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