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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).

John Doe
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2 Answers2

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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.

Anurag A
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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

Mark Bennet
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  • But I still have 3 unknown variables and only two equations? Unless I am missing something. – John Doe May 27 '15 at 21:10
  • @JohnDoe $ax+b=cx^2+dx+e$ is one equation in one unknown because you already know $a,b,c,d,e$. –  May 27 '15 at 21:12
  • @JohnDoe You have two unknowns, because setting the expressions equal eliminates $y$. You have $x$ still as unknown, and what you have called $n$ in your question. But you have two conditions - the first is that $x$ is a root of the quadratic, and the second is that $n$ is such that the quadratic has a single root. These conditions are enough. You need to deploy what you know about quadratics (e.g. complete the square) to reach a solution. – Mark Bennet May 27 '15 at 21:19
  • Yep, I got the right answer. Thanks :) – John Doe May 27 '15 at 21:24
  • No need to complete the square, just set the discriminant to zero. – John Doe May 27 '15 at 21:25
  • @JohnDoe The two are equivalent. Here writing $2x^2+4x+2-n=2(x+1)^2-n$ is an obvious move and you can read off $x_0=-1$ immediately without having to compute $n$ at all, and then get $y_0$ from the quadratic. And of course $n=0$ is obvious too. – Mark Bennet May 27 '15 at 21:29