# equation of the tangent to the following curves?

example: $f(x)=x^2 (2,4)$ $= x^2$
at x= 2
$f(x) 2^2=4$
$y=mx+c$
$y=4x+c$
$(2,4)$
$4=4\cdot2+c$
$4-8=c$
$c=-4$
$y=4x-4$
but below question I don't know how to solve?
$F(x)=x^2-2x+5$ at $x= -1$ the answer is $y=-4x+4$ thanks in advance! :)

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Differentiate $F$ and evaluate at $x=-1$ to get the slope, then use the point-slope form of the equation for a line. –  Guess who it is. Sep 29 '11 at 5:30
can you please explain in a step? thx~ @J.M. –  Sb Sangpi Sep 29 '11 at 5:36
Answer these first: 1. What is the derivative of $x^2-2x+5$? 2. Do you recall how to construct the equation of a line with a given slope and passing through a given point? –  Guess who it is. Sep 29 '11 at 5:38
1. $2x-2$ but i don't know 2? –  Sb Sangpi Sep 29 '11 at 5:50
What you wrote is a bunch of formulas, some of which make no sense, though you arrived at the right tangent line. For instance, there is the $(2,4)=x^2$ in the first line, the $f(x)2^2=4$ on the third line, the bare $(2,4)$ on the sixth line. Connecting text is necessary. We have $f(x)=x^2$. Therefore $f'(x)=2x$. So at $x=2$, the tangent line has slope $f'(2)$, which is $4$. And so on. For the next function, $f(x)=x^2-2x+5$, so $f'(x)=2x-2$. We have $f'(-1)=-4$, so our tangent line has slope $-4$. But $f(-1)=8$, so our tangent line passes through $(-1,8)$. Continue. –  André Nicolas Sep 29 '11 at 6:15

Added: concerning the first part of your question, you should edit it. As a minimum, something like the following.

Example: $f(x)=x^2$. At $x=2$, $f(x)=2^2=4$. The equation of the tangent at $(2,4)$ is

$$y=mx+c,$$

where $m$ is $f'(2)=4$. And so, $y=4x+c$, and $$4=4\cdot 2+c\Leftrightarrow 4-8=c\Leftrightarrow c=-4.$$ Therefore, $y=4x-4$.

Using your notation, the equation of a straight line is

$$y=mx+c\tag{1},$$

where $m$ is its slope. If this line passes through the point $P(a,b)$, then the coordinates $a,b$ must satisfy the equation

$b=ma+c\tag{2}.$

Subtracting $(2)$ from $(1)$, we get

$$y-b=m(x-a),\tag{3}$$

which is equivalent to

$$y=mx-ma+b.\tag{3'}$$

The slope of the tangent line to the graph of the function $f(x)$ at the point $(a,b)=(a,f(a))$ is equal to the derivative of the function at $x=a$, i.e. $m=f^{\prime }(a)$ (Wikipedia link). Therefore, the equation of the tangent is given by (see sketch)

$$y=f^{\prime }(a)x-f^{\prime }(a)a+f(a).\tag{4}$$

For the second function $F(x)=x^{2}-2x+5$, the equation of the tangent at $(-1,F(-1))=(-1,(-1)^{2}-2(-1)+5)=(-1,8)$ is

$$y=F^{\prime }(-1)x-F^{\prime }(-1)\left( -1\right) +8.$$

The derivative $F^{\prime }(x)=2x-2$ and $F^{\prime }(-1)=2\left( -1\right) -2=-4$. Consequently, the equation of the tangent is

$$y=-4x-(-4)\left( -1\right) +8,$$

which is equivalent to

$$y=-4x+4.\tag{5}$$

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(+1) for your generosity, and the (manual) drawings. –  The Chaz 2.0 Sep 29 '11 at 15:04
@TheChaz: Thank you! –  Américo Tavares Sep 29 '11 at 18:17
fantastic answer! Thanks alot. Sorry for my bad question. Now I fully understood~ thx @AméricoTavares –  Sb Sangpi Sep 30 '11 at 2:27
sorry! how do you get $y=-4x-(-4)(-1)+8$ ? –  Sb Sangpi Sep 30 '11 at 2:43
@SbSangpi: Just substitute $F'(-1)=-4$ in the equation $y=F'(-1)x-F'(-1)(-1)+8$. –  Américo Tavares Sep 30 '11 at 9:51