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I have the following task: calculate the coordinates of the intersections point between a straight line with a given slope and a quadratic function, so that you only receive one intersection instead of the normal two or none.
I am given the slope gradient m and the quadratic equation.
In this example its 

y=x^2+3x-2   m=1
user2741831
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  • In order for there to be only one intersection, we need the line to be tangent. Do you have calculus at your disposal. – Sean Haight Nov 22 '16 at 18:38
  • yes I do and I understand most of the quadratic formula material, but I hav etrouble with this one in particular, because I somehow have to force a result where both results of the quadratic formula are identical – user2741831 Nov 22 '16 at 18:40

3 Answers3

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First, find the derivative of $y = x^2 + 3x - 2$.

$$y' = 2x + 3$$

Now set $2x+3 = m = 1$.

Solve for $x$ to find the $x$ coordinate of the point you need.

Then place your solution for $x$ into the equation $y = x^2 + 3x -2$ to find the y-coordinate.

amWhy
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  • I don't really understand where you got y=2x+3 from and where why I should put the gradiant in a as the y coordinate, care to elaborate? – user2741831 Nov 22 '16 at 18:43
  • Have you learned about derivatives? The only way a parabola and a line canintersect at exactly one point, is if $y'(1) = 1$. That is, the line with slope $1$ must be tangent to the parabola when $y'=1 = 2x+3$. – amWhy Nov 22 '16 at 18:45
  • No, I don't think so, but I think I sohuld be able to figure out how they work, I'm not sure if this they way my teacher intended me to solve this, but as long as they work, I fine with this solution, thanks. – user2741831 Nov 22 '16 at 18:49
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Let the equation of straight line be $y = mx +c$. Solve for the intersection: $$ mx+c = x^2 +3x -2$$ You get a quadratic in x. $$ x^2 + (3-m) x -(2+c) = 0$$

To have only one intersection point, there should be only value of $x$ satisfying the above equation. So you make the discriminant of the above equation zero.

So now your x is simply $(m-3)/2$.

You know the value of $m$, so you know $x$, then you can find $y$ to know the intersection point

Swapnil Rustagi
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You don’t really need calculus to solve this problem thanks to a property of the parabola, namely, that a tangent to it doesn’t intersect the parabola at any other points. So, you are essentially looking for values of $b$ such that the line $y=x+b$ and the parabola $y=x^2+3x-2$ intersect in a single point. Setting the right-hand sides equal to each other and rearranging, you get the quadratic equation $$x^2+2x-(b+2)=0.$$ This equation has a single solution when its discriminant is zero, i.e., when $$4-4(b+2)=0.$$ I hope you can take it from here.

amd
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  • just a a question, isnt y=x+b missing the m? So it gives me y=mx+b or y=1x+b?, I know it doesn't acutaly matter this time, but just to make sure. – user2741831 Nov 22 '16 at 18:56
  • @user2741831 We’re given that $m=1$. Swampnil’s answer waits until later to make this substitution. If you want a solution to the general problem, then you need $ax^2+bx+c=mx+k$ to have a single solution for $x$, so you would have to solve $(b-m)^2-4a(c-k)=0$ for $k$. – amd Nov 22 '16 at 19:03
  • Ok, think I got it now – user2741831 Nov 22 '16 at 19:05