Thread: Finding intersection of tangent and line

1. Finding intersection of tangent and line

I have to find the coordinates of the point of intersection of the tangents to the graph of y=x^2 at the points at which it meets the line with equation y=x+2.

This is what I have so far:
y=x^2 intersects with y=x+2 at -1 or 2 since x^2=x+2 factorises to (x+1)(x-2)

The derivative of y=x^2 is 2x , so the gradients of y=x^2 at these points are -2 and 4

I am stuck on where to proceed from here, could somebody please nudge me in the right direction?

Thanks

2. Hello, StaryNight!

Find the coordinates of the point of intersection of the tangents to $\displaystyle y\:=\:x^2$
at the points at which it meets the line $\displaystyle y\:=\:x+2$

The slope of the tangent is given by: .$\displaystyle \frac{dy}{dx} = 2x$

The parabola and line intersect at: .$\displaystyle P(2,4)\,\text{ and }\,Q(\text{-}1,1)$

At $\displaystyle P(2,4)$, the slope is: $\displaystyle m = 4$
The equation of the tangent is: .$\displaystyle y - 4 :=\:4(x-2)\quad\Rightarrow\quad y \:=\:4x-4$

At $\displaystyle Q(\text{-}1,1)$, the slope is: $\displaystyle m = \text{-}2$
The equation of the tangent is: .$\displaystyle y - 1 \:=\:\text{-}2(x+1) \quad\Rightarrow\quad y \:=\:\text{-}2x - 1$

Now find where the two tangents intersect . . .

3. Many thanks Soroban, it turns out I misunderstood the question.

My solution is (0.5,-2)