If the tangent line to y = f (x) at (9,7) passes through the point (0,6)

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Homework Statement


If the tangent line to y = f (x) at (9,7) passes through the point (0,6), find the following.

(a) f (9)

(b) f ' (9)

Homework Equations



f '(x) = f(x+h) - f(x)/h ?

The Attempt at a Solution



I really have no idea how to even begin going about this one, so a shove in the right direction would be greatly appreciated.
 
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my guess:
Since the line is tangent to f(x) at (9,7), then (9,7) must be a point on f(x). If (9,7) is on the function, then when x=9, what is y? That's f(9).

You have 2 points on the tangent line. (9,7) and (0,6). Find the slope of this line using the slope formula: rise / run. Then remember that f'(9) is simply asking what's the slope of the function at (9,7), which is the same as asking what is the slope of the tangent line.
 
Thank you very much! I'm not sure why I had such a hard time conceptualizing what that question was asking, but your hints were correct and I got it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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