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How do you find the equation of the line tangent to the graph of $y = x^{2} - 3$ $y = x^2 - 3$ at the point P(2,1)?

1 Answer

Aug 20, 2015

$4 x - y = 7$ $4x-y=7$

Explanation:

The slope of a tangent to $y = x^{2} - 3$ $y=x^2-3$ is given by its derivative:
$XXXX m = \frac{d y}{d x} = 2 x$ $color(white)("XXXX")m=(dy)/(dx) = 2x$

At $(x, y) = (2, 1)$ $(x,y) = (2,1)$ the slope becomes (substituting $2$ $2$ for $x$ $x$
$XXXX m = 4$ $color(white)("XXXX")m = 4$

The general slope point form for a line with slope $m$ $m$ through a point $(ˆ x, ˆ y)$ $(hatx,haty)$ is
$XXXX y - ˆ y = m (x - ˆ x)$ $color(white)("XXXX")y-haty = m(x-hatx)$

Substituting $m = 4$ $m=4$ , $ˆ x = 2$ $hatx=2$ , and $ˆ y = 1$ $haty=1$
$XXXX y - 1 = 4 (x - 2)$ $color(white)("XXXX")y-1 = 4(x-2)$

This could be re-written in standard form as
$XXXX 4 x - y = 7$ $color(white)("XXXX")4x-y = 7$

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