Find approximation of $y= {x^2}$

Question

I have a function $\large{f(x)=\sqrt{x} \space \space \space x \forall \geq 0}$

I am looking for a quadratic approx. to $f(x)$ at $x=9$.

So far, I know that the quadratic approx. at $x = x_0$ follows the Taylor series:

$$q(x) = f(x_0) + f'(x_0)(x-x_0) + \frac {f''(x_0)}{2} {(x-x_0)}^2$$ and by calculation, $$\begin{align*}f(x) &= 3 \\ f'(x) &= \frac 1 6 \\ f''(x) &= \frac {-1} {108}\end{align*}$$

Are the calculations correct? And happens after this?

Answer 1

So your quadratic is $$q(x) = 3 + \frac{(x-9)}{6} - \frac{{(x-9)}^2}{216}$$

Answer 2

Just to show you how good this approximation is near $x=9$: Here is a graph of $f(x)$ (red), the linear approximation, i.e $y = x+\dfrac16(x-9)$ (blue) and the quadratic approximation $q(x)$ (green):

Answer 3

Let $p(x)=ax^2+bx+c$ for some $a,b,c$ is that polynomial. So we want it to be like: $$p(9)=f(9)\\p'(9)=f'(9)\\p''(9)=f''(9)$$ So we have the following system: $$81a+9b+c=3\\18a+b=1/6\\2a=1/108$$ The same is to solve the system.