Leroy Little

2022-02-12

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

Explanation:
The slope of a tangent to $y={x}^{2}-3$ is given by its derivative:
$m=\frac{dy}{dx}=2x$
At (x,y)=(2,1) the slope becomes (substituting 2 for x) m=4
The general slope point form for a line with slope m through a point
Substituting m=4, $\stackrel{^}{x}=2$, and $\stackrel{^}{y}=1$
y-1=4(x-2)
This could be re-written in standard form as
4x-y=7

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