Find the polynomial that goes through points B and C and is tangent to $f(x) = (2/3)x+4$ at A, Interpolation

Question

I have 3 points:

$$A = (0,4) \\ B=(-5,0) \\ C=(5,0)$$

I need to find a polynomial that goes through B and C, and is tangent to $f(x) = (2/3)x+4$ at A.

I know that tangent means it must be equal to the derivative of f(x) at that point.

This is probably wrong, but I did the interpolation using points B, C and $(0,2/3)$. I got

Help?

Answer 1

You have four conditions $$\begin{align}f(0)&=4\\f(-5)&=0\\f(5)&=0\\f'(0)&=\frac23\end{align} $$ and so we should be looking for four unknowns, i.e., $f$ is of degree $3$, or $$ f(x)=ax^3+bx^2+cx+d$$ and the four conditions translate into linear equations in the unknown coefficients: $$\begin{align}d&=4\\-125a+25b-5c+d&=0\\125a+25b+5c+d&=0\\c&=\frac23\end{align} $$ As $c$ and $d$ are available immediately, you are essentially left with a simple system of two very user-friendly linear equations in $a$ and $b$.

Your interpolation considers only the first three givens.

Answer 2

$P(x)=(x^2-25)(ax+b)$

$P(0)=-25b=4\iff b=\dfrac{-4}{25}$

$P'(x)=3ax^2-25a-\dfrac{-4x}{25}\Rightarrow P'(0)=-25a=\dfrac23\Rightarrow a=\dfrac{-2}{75}$