If a line parallel to $y=-7x+3$ touches the parabola $2x^2-3x+2$ in the point $(x_0,y_0)$, what is the value of $4x_0+y_0$?

Question

I tried solving this but I've no idea how to find the point where a line of the form $y=-7x+n$ intersects a given parabola.

Hints are welcome. Don't know calculus (don't even know if it's applicable here, but either way).

Answer 1

Hint: the line $y=ax+b$ meets the parabola $y=cx^2+dx+e$ when $ax+b=cx^2+dx+e$ - can you see why?

Second hint: the tangent to a parabola meets it at just one point as you will see from a simple sketch

Answer 2

The slope of the tangent line to the parabola at $(x_0,y_0)$ is given by $4x_0-3$, but this should be $-7$. Thus $x_0=-1$. This gives $y_0=2(-1)^2-3(-1)+2=7$. Now you can compute $4x_0+y_0$.

I didn't read your comment about calculus: so here is one without calculus:

Let the tangent line be $y=-7x+b$, then the intersection with parabola can be computed as: $$-7x+b=2x^2-3x+2.$$ This leads to $$2x^2+4x+(2-b)=0.$$ For tangency it should have equal roots, thus discriminant $4^2-8(2-b)=0$. Now you get your $b$ and you can compute $(x_0,y_0)$ from here.