Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?!

share|cite|improve this question
Have you tried using the hint given? – J. M. 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:

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

share|cite|improve this answer
pedja, for things like these, it's usually not a good idea to give full solutions... – J. M. 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. 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 $y-2=\frac{-1}4(x-2)$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.