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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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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. – t.b. May 21 '11 at 15:15
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? – aka May 21 '11 at 15:21
@aka: It is solvable, using tools that you possess. – André Nicolas May 21 '11 at 15:37
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? – aka May 21 '11 at 15:48
Many thanks for your help, Américo and Chandru.... I got it right know! Had even forgotten the equation of tangential! Once again thank you! – aka May 21 '11 at 19:05

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? – aka May 21 '11 at 16:01
@aka: Please note that as you decrease the value of $a$ the parabola becomes closer to the line $y=2x+3$ – user9413 May 21 '11 at 16:04
Meaning smaller a is, the closer to our function y = 2x + 3 – aka May 21 '11 at 16:09
@aka: Also, i think you need a point at which the line $2x+3$ is tangential to the given parabola. – user9413 May 21 '11 at 16:10
@aka: Exactly. I shall add another picture. – user9413 May 21 '11 at 16:10

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) + x – aka May 21 '11 at 16:06
Except $a$ is not a function in terms of $x$; it is a number. – Hans Parshall May 21 '11 at 16:13
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. – Hans Parshall May 21 '11 at 16:14
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 $(c, 2c+3)$. – André Nicolas May 21 '11 at 16:19
@aka: I plotted here the graph corresponding to Hans Parshall's solution… (If there is any error, the fault is mine). – Américo Tavares May 21 '11 at 16:28

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


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… to the question "Find equation of quadratic when given tangents?"… – Américo Tavares May 21 '11 at 17:47
Using your's I got my algebraic solution. Thanks – user9413 May 21 '11 at 18:24
@Chandru: You are welcome! – Américo Tavares May 21 '11 at 18:31
Oh, No problem. +1 for your answer. – user9413 May 21 '11 at 18:32

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