Quadratic approximation of $9^{1/3}$ Find a quadratic approximation of the cube root of $9$ by using the equation $9=8(1+\frac 18)$, and estimate the difference between the exact value and the approximation.
How am I supposed to find the quadratic approximation of this formula? If there is no $x$, and the formula for quad approx is $Qa(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2$, so does it mean that my quadratic approx will be equal just to $f(a)$?
Could anyone help please, it looks easy, but I can't get it. 

When you write $$P(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2,$$this polynomial is the approximation near $a$. This means that $P(x)$ will be close to $f(x)$, if $x$ is close to $a$. Another way to express this formula is writing $x - a = h$. So: $$P(a+h) = f(a) + f'(a) h + \frac{f''(a)}{2}h^2.$$This expresses the approximation in terms of the difference $h$, and the closer $h$ is to zero, closer $P(a+h)$ will be to $P(a)$. 
Here your intuition should be clear: we want to approximate $\sqrt[3]{9}$, but the closest integer cube root we know is $\sqrt[3]{8} = 2$. So we want to approximate $\sqrt[3]{x}$ near $8$ ($a = 8$). We want the value of $9$ ($x = 9$). The difference is $1$ ($h = 1$). Now you can use any of the expressions above (pick your favorite).
