Find the equation of the tangent line to f(x)=sqrtx^2+6x at x=2, give your answer in slope-intercept form

Answer 1

can't tell if you mean
f(x) = √(x^2) + 6x
or
f(x) = √(x^2 + 6x)

I will assume the latter
f'(x) = (1/2)(2x)/√(x^2 + 6x)
f'(2) = 2/√16 = 1/2

also f(2) = .....

So now you have the slope and a point, you must be able to finish it

Answer 2

I meant f(x)sqrt(x^2+6x), thank you!

Answer 3

Am I doing this correct?

y-y1=m(x-x1)
y-1/2=-1/2(x-2)
y-1/2=-1/2x+1
add 1/2 to both sides
y=-1/2x+1.5

Answer 4

Your y value of the point is not correct

f(2) = √(2^2 + 6(2)) = √ 16 = 4

so the point is (2,4) and the slope is 1/2

y-4 = (1/2)(x-2)
y = (1/2)x - 1 + 3
y = (1/2)x + 2