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! :)
equation of the tangent to the following curves?
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2Differentiate $F$ and evaluate at $x=-1$ to get the slope, then use the point-slope form of the equation for a line. – 2011-09-29
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0can you please explain in a step? thx~ @J.M. – 2011-09-29
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1Answer 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? – 2011-09-29
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01. $2x-2$ but i don't know 2? – 2011-09-29
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5What 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. – 2011-09-29
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0If the line $y=mx+c$ passes through the point $(a,b)$ then $b=ma+c$ so $c=b-ma$. – 2011-09-29
1 Answers
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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0fantastic answer! Thanks alot. Sorry for my bad question. Now I fully understood~ thx @AméricoTavares – 2011-09-30
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0sorry! how do you get $y=-4x-(-4)(-1)+8$ ? – 2011-09-30
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0@SbSangpi: Just substitute $F'(-1)=-4$ in the equation $y=F'(-1)x-F'(-1)(-1)+8$. – 2011-09-30