f(x) = (3x+5)^.5
f'(x) = .5 (3x+5)^-.5 * 3
= 1.5 (3x+5)^-.5
when x = 10
f(x) = sqrt(35)
f'(x) = 1.5 /sqrt(35)
y = m x + b
sqrt(35) = [1.5/sqrt(35)]10 + b
35 = 15 + b(sqrt(35))
20 /sqrt(35) = b
so
y = [1.5/sqrt(35)] x + 20 /sqrt(35)
or
sqrt(35) y = 1.5 x + 20
equation of tangent line of the function sqrt (3x+5) where x=10
3 answers
thanks Damon, in HS and this is kicking my butt!
You are welcome.