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I am stuck with a problem I simply cannot solve.

I have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:

The three lines described by the equations

$y_1(x)=-4x-16.5$

$y_2(x)=2x-4.5$

$y_3(x)=6x-16.5$

are all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$

Determine the values of the coefficients a, b & c.

I simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)

Edit: I'm including the way I tried to solve it. I didn't get super far.

Given the polynomial $p(x)$ I know that $p'(x)=2ax+b$

Therefore, the following is true for the three points with x-values of $x_1, x_2 $ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:

$-4=2ax_1+b$

$2=2ax_2+b$

$6=2ax_3+b$

That's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.

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    $\begingroup$ Show anyway what you tried. I can give a solution using derivatives, but maybe it is not the solution you want. $\endgroup$ – N74 May 8 '18 at 20:23
  • $\begingroup$ Welcome to SE! Even if your attempts didn't work, you should absolutely include them in your post - people can advise you better that way! $\endgroup$ – B. Mehta May 8 '18 at 20:23
  • $\begingroup$ @N74 I have now included my failed solution :) As you can see, I thought of using derivatives as well, but I can't get further. $\endgroup$ – Mads Clausen May 8 '18 at 20:38
  • $\begingroup$ My maths teacher (I'm in the last year of Danish "high school") gave it to us for tomorrow. He said that it was definitely doable. $\endgroup$ – Mads Clausen May 8 '18 at 20:42
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If we have a parabola $$ y = a x^2 + b x + c $$ and a line $$ y = mx + d,$$ they are tangent if $$ b^2 - 4ac = -4da + 2mb - m^2 $$ as then the parabola $$ y = a x^2 + (b-m) x + (c -d) $$ has a double root, i.e. is a constant times a square, $$ a (x+p)^2 $$ Might as well write this: I fixed $$ \Delta = b^2 - 4 a c $$ and then had three equations $$ -4da + 2mb = m^2 + \Delta $$ by plugging in the values from the three lines $ y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\Delta,$ after which we have one equation for $c$

$$ 66a - 8 b = 16 + \Delta \; , $$ $$ 66a + 12 b = 36 + \Delta \; , $$ $$ 18a + 4 b = 4 + \Delta \; . $$ Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\Delta.$ Finally, $c = \frac{b^2 - \Delta}{4a}$

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Note that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero \begin{eqnarray*} ax^2+(b-m)x+c-d=0. \end{eqnarray*} This will lead to $3$ equations for $a,b,c$ that are easily solved giving

$(a,b,c)=(1/2,1,-4)$.

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  • $\begingroup$ Thanks for the answer. I don't quite follow, however: what are the three equations for a, b and c that you mention? $\endgroup$ – Mads Clausen May 8 '18 at 21:02
  • $\begingroup$ \begin{eqnarray*} (b+4)^2=2a(c+33) \\ (b-2)^2=2a(c+9) \\ (b-6)^2=2a(c+33) \end{eqnarray*} combine the first & the third of these equations & you will rapidly yield a value for $b$. $\endgroup$ – Donald Splutterwit May 8 '18 at 21:05
  • $\begingroup$ Donald I get different final abc values, although still all rational and easily found... I kept the discriminant $\Delta = b^2 - 4ac \; $ as an unknown until the end, began with three $ \Delta = -4da + 2mb - m^2 $ or $-4da + 2mb = m^2 + \Delta $ $\endgroup$ – Will Jagy May 8 '18 at 21:09
  • $\begingroup$ where does $2a(c+n)$ come from? I recall that the discriminant is equal to $b^2-4ac$, which in this case would yield something like $(b-m)^2=4a(c-d)$ I would think it would either be that or $(b-m)^2=2a(2c-2d)$ $\endgroup$ – Mads Clausen May 8 '18 at 21:12
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    $\begingroup$ I get 1/2, 1 and -4 for a,b,c. 2ax+b = -4 at x=-5; 2 at x = 1 and 6 at x = 5 with x and y values being the same for tangent lines and parabola at these tangent points. $\endgroup$ – Phil H May 8 '18 at 21:32
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Given any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.

Working out the intersection points of the three lines, we can see that the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$ and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$ Let

  • $x = x_1$ at the tangent point with $y = y_1(x)$;
  • $x = x_2$ the tangent point with $y = y_2(x)$; and
  • $x = x_3$ the tangent point with $y = y_3(x).$

Due to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$; moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$ It follows that $$x_3 - x_1 = 2\left(\frac{x_3 + x_2}2 - \frac{x_2 + x_1}2\right) = 2(3 - (-2)) = 10.$$

The tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and $f'(x_3) = y_3'(x) = 6.$ But $f'(x) = 2ax + b,$ so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ and therefore $a = \frac12.$ It is then a relatively straightforward exercise to find the other two coefficients.

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