f' = -27(2x)/(x^2+9)^2
at x = -3
f'=slope = -27(-6)/(18)^2 = .5
so
y = .5 x + b
put in the point
1.5 = .5 (-3) + b
1.5 = -1.5 + b
b = 3
so
y = .5 x + 3
or
2y = x + 6
Find an equation of the tangent line to the graph of f at the given point.
f(x)=27/(x^2+9)
(-3,3/2)
I got y=x/2+3, but I don't think its right.. please help!
1 answer
1 answer