I have the function $$y=2x^2-3x$$ How do I find the co-ordinates of the point of intersection of the lines tangent to the curve at $y=-1$?
One point where $y = -1$ is when $x = \dfrac 12$.
I took the derivative of the given $y$: $\dfrac{dy}{dx} = 4x- 3$, then found slope at $x = \dfrac 12$. So slope of the tangent line is $-1$.
So the equation of tangent is $y+1=-1\left(x-\dfrac 12\right)$.
How to solve it further?
Solve for $x$ when $y = -1$: $$2x^2 - 3x = -1 \iff 2x^2 - 3x + 1 = 0 \iff (2x-1)(x-1) = 0 \implies x = \frac 12 \text{ or } x = 1$$
So, there are two points corresponding to $y = -1$: $\left(\frac 12, -1\right), (1, - 1)$.
Perhaps you need to find the point of intersection of the line tangent to the first point, with the line tangent to the second point.
Indeed, you've found the equation of the line tangent to the point $\left(\frac 12, -1\right)$, since $-1 = \dfrac{dy}{dx}\left(\frac 12\right)$: $$y+1= -\left(x -\frac 12\right)\iff y = -x -\frac 12\tag{1}$$
Now we need to find the equation of the line tangent to $(1, -1)$. $$\frac{dy}{dx}(1) = 4(1)-3 = 1$$
So the equation of the second line is given by $$y +1 = x-1 \iff y = x-2\tag{2}$$
Now, put $y$ from $(1)$ equal to $y$ from $(2)$, and solve for $x$, the x coordinate of the point of intersection. Then solve for $y$ using the equation for either line.