5
$\begingroup$

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

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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    What have you tried to do? Can you turn the two conditions on the function $y = ax^2 + bx + c$ into conditions on $a,b,c$? The first condition determines $b$ and $c$. Now you only have $a$ left to figure out, and for that you'll need the second condition.2011-05-21
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    Yes, I've got b and c. b = 1 and c = 0 a = pain in my ass? We need a second condition to solve it right?2011-05-21
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    @aka: It is solvable, using tools that you possess.2011-05-21
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    How? so far I got the function: y = ax^2 + x and we know that it tangetial with the line y = 2x + 3, thus in which point we dont know. So if i put these to against eachother ax^2 + x = 2x + 3 we get: a = (3/ X^2-x) So our function becomes y = (3x^2)/(x^2-x) + x Is this right?2011-05-21
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    Many thanks for your help, Américo and Chandru.... I got it right know! Had even forgotten the equation of tangential! Once again thank you!2011-05-21
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    @aka: please do not use answers to leave comments.2011-05-21
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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

4

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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    Who should this help? and how did you get -1/5 ? Could you show me step by step how solve or give me a hint?2011-05-21
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    @aka: Please note that as you decrease the value of $a$ the parabola becomes closer to the line $y=2x+3$2011-05-21
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    Meaning smaller a is, the closer to our function y = 2x + 32011-05-21
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    @aka: Also, i think you need a point at which the line $2x+3$ is tangential to the given parabola.2011-05-21
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    @aka: Exactly. I shall add another picture.2011-05-21
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    Thank you in advance... look at my comment at the answer... is that right ?2011-05-21
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    @aka: Picture added2011-05-21
3

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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    That you are saying is that in one point x, the two functions are equal! Which gives us (after a bit math) a = 3/ (X^2-x) there for our funktion will be: y = (3x^2)/(x^2-x) + x2011-05-21
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    Except $a$ is not a function in terms of $x$; it is a *number*.2011-05-21
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    The one point $x_0$ where the two functions are equal is also a number; you need to solve for both this and $a$ to complete the problem.2011-05-21
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    In my experience, it is dangerous for students to use $x$-like notation for the point of tangency with $y=2x+3$. The probability of success is greater if we let the point of tangency be $(c, 2c+3)$.2011-05-21
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    @aka: I plotted here the graph corresponding to Hans Parshall's solution http://calcauxprobteor.files.wordpress.com/2011/05/quadratic2tangents2.jpg?w=200&h=300 (If there is any error, the fault is mine).2011-05-21
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    @user6312: I had the same fear, which is why I included the subscript, but this was still too close to $x$. I'll edit my answer in line with your suggestion.2011-05-21
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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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    This method is very similar to this answer of mine http://math.stackexchange.com/questions/31321/find-equation-of-quadratic-when-given-tangents/31381#31381 to the question "Find equation of quadratic when given tangents?" http://math.stackexchange.com/questions/31321/find-equation-of-quadratic-when-given-tangents2011-05-21
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    Using your's I got my algebraic solution. Thanks2011-05-21
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    @Chandru: You are welcome!2011-05-21
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    Oh, No problem. +1 for your answer.2011-05-21