# Thread: Equation of a Tangent

1. ## Equation of a Tangent

The tangent to the curve $\displaystyle y = px^3$ at the point A where $\displaystyle x=2$, meets the y-axis at B(0,8). Find the gradient of the curve at point A.

I have attempted using differentiation and found the gradient to be 12p... However, gradient must be in value form..

2. Originally Posted by Punch
The tangent to the curve $\displaystyle y = px^3$ at the point A where $\displaystyle x=2$, meets the y-axis at B(0,8). Find the gradient of the curve at point A.

I have attempted using differentiation and found the gradient to be 12p... However, gradient must be in value form..
hi

At x=2 , y=p(2)^3=8p so the tangent touches the curve at (2,8p)

gradient of tangent = (8p-8)/2

so 8p-8/2=12p ..

3. I already did this in my attempt... The gradient MUST be in value form.

4. Originally Posted by Punch
I already did this in my attempt... The gradient MUST be in value form.
look closer , $\displaystyle \frac{8p-8}{2}=12p$ , solving this , p=-1/2 so

dy/dx=12p=12(-1/2)=-6