1
$\begingroup$

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! :)

  • 0
    If the line $y=mx+c$ passes through the point $(a,b)$ then $b=ma+c$ so $c=b-ma$.2011-09-29

1 Answers 1

6

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)

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

enter image description here

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

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

  • 0
    @SbSangpi: Just substitute $F'(-1)=-4$ in the equation $y=F'(-1)x-F'(-1)(-1)+8$.2011-09-30