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How would I solve the following?

Find an equation for the tangent line to the graph

$f(x)=3x^2-4x$ at the point $x=-1$

What I did is

$3(-1+h)^2-4(-1+h)-7)$

$3-6h+3h^2+4-4h-7)$

As h approaches zero $(2h+3h^2)/h$ limit equal $2$

then the equation I wrote is $y-7=2(x+1)$

But I am not sure if it is correct.

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Very nice use of LaTeX (MathJax)! +1 –  Thomas Jan 31 '13 at 22:20
    
@thomas thanks. –  Fernando Martinez Jan 31 '13 at 22:22
    
don't you know derivatives? –  Emanuele Paolini Jan 31 '13 at 22:23
2  
@manu-fatto - don't be so presumptuous: there are steps one takes in learning the definition of the derivative, using limits, before learning direct differentiation. I find your comment to be demeaning. Certainly, it is not at all helpful here. –  amWhy Jan 31 '13 at 22:34
    
May I suggest simplifying the equation into either standard form 2x-y=-9 or the slope-intercept form of y = 2x+9 ? –  JB King Jan 31 '13 at 22:38
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2 Answers

up vote 6 down vote accepted

You're approach is sound: the one (simple algebraic) mistake is that

$$(3-6h+3h^2+4-4h-7)\, =\, 3h^2 - 10h,\; \text{ and not} \;\;3h^2 + 2h:$$


So we need to evaluate the limit of $\;\;\dfrac{(3h^2 -10h)}{h}\; = \;\dfrac{h(3h -10)}{h}\;$ as $\;h\to 0:$

$$ \lim_{h\to 0}\,\frac{h(3h - 10)}{h} \; =\;\lim_{h\to 0}\, (3h-10) = -10.$$


So you need only replace "$2$" with "$-10$" in your equation of the line tangent to $\;f(x)=3x^2-4x\;$ at the point $\;(-1, 7)\;$, which gives you:

$$y - 7 = {-10}(x +1) \;\iff \;y = -10x-3$$

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Thanks for the answer. –  Fernando Martinez Jan 31 '13 at 22:36
    
You're very welcome: nice work! and nice formatting! Your ONLY mistake was in the computing the coefficient of the "h" term. You handled the conceptual material perfectly. –  amWhy Jan 31 '13 at 22:38
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So you are finding the limit for the slope $$\begin{align} f'(-1) &= \lim_{h\to 0} \frac{f(-1 + h) - f(-1)}{h} \\ &= \lim_{h\to 0} \frac{3(-1+h)^2 + 4 - 4h - 7}{h}\\ &= \lim_{h\to 0}\frac{3 - 6h + 3h^2 -4h - 3}{h}\\ &=\lim_{h\to 0}\frac{h(-10 + 3h)}{h}\\ &=\lim_{h\to 0} -10 + 3h = -10. \end{align} $$

And the indeed an equation of the tangent line is $$ y - 7 = {\color{red}{-10}}(x +1) $$

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I see thanks for your answer. –  Fernando Martinez Jan 31 '13 at 22:36
    
@FernandoMartinez: Glad to help. –  Thomas Jan 31 '13 at 22:38
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