# Slope of the tangent lines?

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?!

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Have you tried using the hint given? –  J. M. is back. Oct 8 '11 at 5:55
Do you know how to take the derivative of $4x^2 + 6x + 7$? –  Unreasonable Sin Oct 8 '11 at 6:06

Equation of tangent line at point $(a,f(a))$ is $y = f(a) + f'(a)(x - a)$, so we have to find $f'(x)$ and than plug in value $a$ into the result.

$f'(x)=(4x^2+6x+7)'=8x+6 \Rightarrow f'(3)=30$

Since $f(a)=f(3)=61$, we may write next tangent line equation:

$y=61+30(x-3) \Rightarrow y=30x-29$

For the second tangent line we have that $f'(x)=\frac{-1}{2\sqrt{10-x}}\Rightarrow f'(6)=\frac{-1}{4}$ ,and $f(6)=2$, so the second tangent line is:

$y=2-\frac{1}{4}(x-6) \Rightarrow y=\frac{-1}{4}x+\frac{7}{2}$

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pedja, for things like these, it's usually not a good idea to give full solutions... –  J. M. is back. Oct 8 '11 at 6:30
@J.M,Do you think that full solution is counterproductive ? –  pedja Oct 8 '11 at 6:37
Ideally we want to leave students something they have to figure out on their own... –  J. M. is back. Oct 8 '11 at 6:39

Let find $m_{\text{tangent}}=f'(6)$ when $f'(x)=\sqrt{10-x}$ by using the definition not applying rules. We know that $$f'(6)=\lim_{x\to 6}\frac{f(x)-f(6)}{x-6}=\lim_{x\to 6}\frac{\sqrt{10-x}-2}{x-6}$$ which is $\frac{0}{0}$. So we do it as follows: $$\lim_{x\to 6}\frac{\sqrt{10-x}-2}{x-6}=\lim_{x\to 6}\frac{(\sqrt{10-x}-2)\color{blue}{(\sqrt{10-x}+2)}}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}=\lim_{x\to 6}\frac{(10-x)-4}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}\\ =\lim_{x\to 6}\frac{6-x}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}=\frac{-1}{4}$$ Now the tangent line at $(6,2)$ is $y-2=\frac{-1}4(x-2)$.

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If you do not know how to take the derivative, you could try computing for the slope using $m = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$

All you have to do is to plug in the values: $$lim_{\Delta x \to 0}\frac{4(x + \Delta x)^2 + 6(x + \Delta x) + 7 - (4x^2 + 6x + 7)}{\Delta x}$$ and you should get $\lim_{\Delta x\to0} 8x + 4\Delta x + 6$

Also, review the equation of a line: $y-y_1 = m(x - x_1)$

Voila! You should now have a tangent line.

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