# What are a , b and c?

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

which is tangent at the beginning with the line $y=x$, It is additionally digressive with the line $y=2x + 3$. Establish the function! Attract a number!

My major inquiry is this understandable? I am uncertain?

5
2022-06-07 14:30:31
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Your trouble is since you have $y = ax^2 + x$ digressive to $y = 2x + 3$. This suggests you have some number $n$ where $an^2 + n = 2n + 3$ (they fulfill at a factor) and also $2an + 1 = 2$ (they fulfill tangentially).

You have 2 formulas and also 2 unknowns ; I make certain you can address from below.

3
2022-06-07 14:56:37
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This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed making use of KmPlot. The number needs to offer you an instinctive suggestion of just how to deal with addressing.

• The Green line is $y=2x+3$.

• Heaven line is $y=x$.

If the line $y=2x+3$ and also the parabola $y=ax^{2}+bx+c$ are mosting likely to be tangent at an offered factor after that their inclines are equivalent. Allow is locate that out. Incline of line $y=2x+3$ is $2$ and also we have $$2 = \frac{dy}{dx} = 2ax+1$$ So you have $x=\frac{1}{2a}$. Additionally we have \begin{align*} 2x+3 & = ax^{2} + x \end{align*} which claims 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 $a=-\frac{1}{7}$.

4
2022-06-07 14:55:09
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Note : The method listed below is really comparable to my response to the inquiry "Find equation of quadratic when given tangents?".

Given that the by-product of $y=f(x)=ax^{2}+bx+c$ is $f^{\prime }(x)=2ax+b$, the formulas of the tangents to the chart of $f(x)$ at factors $(x_{i},f(x_{i}))$, 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 factors is $(x_{1},f(x_{1}))=(0,0)$. As the formula of the tangent at $(0,0)$ is $y=x$ we have to have

$$bx+c\equiv x.$$

Comparing coefficients we get $b=1,c=0$. Therefore $f(x)=ax^{2}+x$. In a similar way for the tangent at $(x_{2},f(x_{2}))$ we have to additionally have

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

Comparing once more coefficients, we get the adhering to system in $a$ and also $x_2$, which allows us to locate $a$:

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

From the first formula we get $x_{2}=1/(2a)$, which by replacement in the 2nd formula offers $a=-1/12$.

Consequently the square formula $y=f(x)$ is

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

Below is the chart of $y=f(x)$ along with its 2 tangents at factors $(0,0)$ and also $(-6,-9)$.

3
2022-06-07 14:54:01
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