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I am learning about Linear and Quadratic approximation.

Quadratic Approximation Formula:

$Q(x) = f(a)+f'(a)(x-a)+\frac{f"(a)}{2}(x-a)^2$

Linear Approximation Formula:

$f(x) = f(a)+f'(a)(x-a)$

I am being told to sketch the graph of $f(x) = (9+x)^2$, as well as the linear and quadratic approximations of $f(x)$ at x = 0.

I am confused how to find the "a" in the formula. But here is what I do know, with x = 0:

$f(0) = 81$

$f'(x) = 2(9+x)$

$f'(0) = 18$

$f''(x) = 2$

$f''(0) = 2$

But, I can't plug this into the formula, because I didn't find "a", right?

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    $\begingroup$ First fix $a$ (for example $a=0$) Then, the formula approximates $f(x)$ in the neighborhood of $a$. In your exercise, we have $a=0$, so you are on the right track. $\endgroup$ – Peter Dec 1 '16 at 22:19
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    $\begingroup$ The $a$ is the point near which the approximation is made. In your case this is $0$. $\endgroup$ – Ian Dec 1 '16 at 22:20
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    $\begingroup$ Tip: whenever the exercise says "near $x=\text{__}$" or "at $x=\text{__}$" or "with $x=\text{__}$", the $\text{__}$ is your $a$. $\endgroup$ – user137731 Dec 1 '16 at 22:30
  • $\begingroup$ So, in the formula, the (x-a) part would be (0-0)?... $\endgroup$ – LaSpana101 Dec 1 '16 at 22:57
  • $\begingroup$ No. In the formula, $x$ is your independent variable -- it just stays $x$. But the $a$ is $0$. $\endgroup$ – user137731 Dec 3 '16 at 16:04

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