5
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$y = ax^2 + bx + c$

which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!

My main question is this solvable? I am doubtful?

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    @aka: Please accept an answer, so that the question is not in the unanswered list.2011-05-23

3 Answers 3

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This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed using KmPlot. The figure should give you an intuitive idea of how to go about solving.

  • The Green line is $y=2x+3$.

  • The Blue line is $y=x$.

If the line $y=2x+3$ and the parabola $y=ax^{2}+bx+c$ are going to be tangent at a given point then their slopes are equal. Let's find that out. Slope of line $y=2x+3$ is $2$ and we have $2 = \frac{dy}{dx} = 2ax+1$ So you have $x=\frac{1}{2a}$. Also we have \begin{align*} 2x+3 & = ax^{2} + x \end{align*} which says that $2 \times \frac{1}{2a} + 3 = a \times \frac{1}{4a^{2}} + \frac{1}{2a}=\frac{3}{4a}$ From this we have $\frac{1}{a} -\frac{3}{4a} = -3 \Longrightarrow a=-\frac{1}{12}$

enter image description here

This is for the value $a=-\frac{1}{3}$ enter image description here

This is for the value $a=-\frac{1}{7}$. enter image description here

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    @aka: Picture added2011-05-21
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Your problem is now that you have $y = ax^2 + x$ tangential to $y = 2x + 3$. This means you have some number $n$ where $an^2 + n = 2n + 3$ (they meet at a point) and $2an + 1 = 2$ (they meet tangentially).

You have two equations and two unknowns; I'm sure you can solve from here.

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    @aka: I get$a$different result for $a$ in terms of $n$ (which is the $x$ in your comment). The easiest substitution I found wat to take $2an+1=2$ to $an=\frac{1}{2}$ and substitute that into the other equation. I wound up agreeing with Americo Tavares' solution.2011-05-21
3

Note: The method below is very similar to my answer to the question "Find equation of quadratic when given tangents?".

Since the derivative of $y=f(x)=ax^{2}+bx+c$ is $f^{\prime }(x)=2ax+b$, the equations of the tangents to the graph of $f(x)$ at points $(x_{i},f(x_{i}))$, with $i=1,2$ are

$\begin{eqnarray*} y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i}) \\ &=&\left( 2ax_{i}+b\right) x-\left( 2ax_{i}+b\right) x_{i}+ax_{i}^{2}+bx_{i}+c. \end{eqnarray*}$

One of the points is $(x_{1},f(x_{1}))=(0,0)$. As the equation of the tangent at $(0,0)$ is $y=x$ we must have

$bx+c\equiv x.$

Comparing coefficients we get $b=1,c=0$. Hence $f(x)=ax^{2}+x$. Similarly for the tangent at $(x_{2},f(x_{2}))$ we must also have

$\left( 2ax_{2}+1\right) x-\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}\equiv 2x+3.$

Comparing again coefficients, we get the following system in $a$ and $x_2$, which enables us to find $a$:

$\left\{ \begin{array}{c} 2ax_{2}+1=2\qquad\qquad\qquad \\ -\left( 2ax_{2}+1\right) x_{2}+ax_{2}^{2}+x_{2}=3.% \end{array}% \right. $

From the first equation we get $x_{2}=1/(2a)$, which by substitution in the second equation gives $a=-1/12$.

Therefore the quadratic equation $y=f(x)$ is

$y=-\frac{1}{12}x^{2}+x.$

Below is the graph of $y=f(x)$ together with its two tangents at points $(0,0)$ and $(-6,-9)$.

enter image description here

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    Oh, No problem. +1 for your answer.2011-05-21