What are a , b and c?

Question

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

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?

Answer 1

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 $$\frac{1}{a} -\frac{3}{4a} = -3 \Longrightarrow a=-\frac{1}{12}$$

This is for the value $a=-\frac{1}{3}$

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

Answer 2

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.

Answer 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 $(-6,-9)$.