Can you help me with this tangent line problem?

Question

 How do I write the slope-intercept form of the equation of the line tangent to the graph y=3x^2+6x+1?

Answer 1

You need to state the point where the line is tangent.  This essential information was not included in the problem statement.  Get the slope of the tangent line from

 

f'(x) = 6x + 6

 

Suppose the line is tangent at (x₁,y₁).  Then

 

m = f'(x₁) = 6x₁ + 6 = 6(x₁ + 1)

 

In point-slope form, the equation of the line is

 

y - y₁ = m(x - x₁)

 

If you want it in the form 

 

y = mx + b,

 

then

 

y = mx - mx₁ + y₁

 

y = mx + (y₁ - mx₁),

 

where

 

b = y₁ - mx₁

 

m = 6(x₁ + 1)