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The slope of the tangent line to the parabola $y = x^2 + 6x + 7$ at the point $( 3 , 61 )$ is: The equation of this tangent line can be written in the form $y = mx+b$ where $m$ is: __ and where $b$ is: __?

and also:

Let $f(x) = \sqrt{10-x}$ The slope of the tangent line to the graph of $f(x)$ at the point $(6,2)$ is . The equation of the tangent line to the graph of $f(x)$ at $(6,2)$ is $y=mx+b$ for $m=$ __ and $b=$ __ Hint: the slope is given by the derivative at $x=6$, ie. $\lim_{x\to6} \frac{(f(6+h)-f(6))}{h}$

I'm absolutely stumped.... :( Help?!

  • 1
    Have you tried using the hint given?2011-10-08
  • 0
    Do you know how to take the derivative of $4x^2 + 6x + 7$?2011-10-08

3 Answers 3