Find an equation of the tangent line to a graph

Question

Consider the graph defined by

$$y = \dfrac{x - 3}{x - 5}$$

(a) Use the definition of the derivative to find the slope of the tangent line to the graph at the point $(6, 3)$.

(b) Find an equation of the tangent line to the graph at the point $(6, 3)$.

Related : http://math.stackexchange.com/questions/45336/relation-between-the-tangent-to-a-curve-and-the-first-derivative-of-a-function — lab bhattacharjee, May 25 '13 at 13:35

score 2 · Answer 1 · answered May 25 '13 at 13:34

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The definition of the derivative , which gives the slope is $$y_0'=\lim\limits_{x\to x_0}\dfrac{y-y_0}{x-x_0}$$ The equation of the tangent line is $$y-y_0=y'_0(x-x_0)$$ Can you now continue?

answered May 25 '13 at 13:34

Dimitris

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I thank you for your help.Using this definition I find the slope of the tangent line to be 0.4 and the resulting equation to be y = 0.4x + 0.6 – Tangles May 25 '13 at 13:43

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

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$$\dfrac {dy}{dx}=\dfrac{(x-5)\dfrac {d}{dx}(x-3)-(x-3)\dfrac {d}{dx}(x-5)}{(x-5)^2}$$

$$\dfrac {dy}{dx}=\dfrac{(x-5)-(x-3)}{(x-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(x-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(6-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=-2$$

slope of tangent at $(6,3)$=$-2$

b)

equation of tangent : $y-y_0=m(x-x_0)$

(where $(x_0,y_0)$ is a point on line and m is slope)

$$y-3=-2(x-6)$$ $$2x+y=15$$

edited Jun 12 '20 at 10:38

Community

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answered May 25 '13 at 13:39

iostream007

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Thankyou so much :) Rockin answer! – Tangles May 25 '13 at 13:49
@Tangles my pleasure.Nothing rocking Just an starter question. – iostream007 May 25 '13 at 13:51
what can I say.... I'm a freshie in the world of calculus ;P – Tangles May 25 '13 at 13:53
keep continue buddy – iostream007 May 25 '13 at 13:54
This can also be greatly simplified by the observation that $\frac{x-3}{x-5}=1+\frac{2}{x-5}$. – rurouniwallace May 25 '13 at 13:59
@ZettaSuro but using differentiation is standard way to do so. – iostream007 May 25 '13 at 14:07
Yes. And taking the derivative of the above expression is much easier, is what I'm saying. – rurouniwallace May 25 '13 at 14:09
okay I got it.that's why downvote. – iostream007 May 25 '13 at 14:10
No, I'm not the one who downvoted, your answer is still correct so I'm not sure what that was for. – rurouniwallace May 25 '13 at 14:21

Find an equation of the tangent line to a graph

2 Answers2

b)