Find the equation of the line tangent to y = –x2 + 3x + 8 at x = 2.

Take derivitive

y'=-2x+3

at x=2
y'=-2(2)+3
y'=-4+3
y'=-1

slope is -1

find the value of y when x=2

y=-(2)^2+3(2)+8
y=-4+6+8
y=10

(2,10) and slope is -1

use point slope form
y-y1=m(x-x1)
a point is (x1,y1) and the slope is m

so
y-10=-1(x-2)
y-10=-x+2
y=-x+12

2nd option

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