# Let f(x)= 6x+15-3e^x. Then the equation of the tangent line to the graph of f(x) at the point (0,12) is given by y=mx+b. Find m and b

### 1 Answer | Add Yours

You need to remember that the equation of the tangent line to the graph of the function at a certain point is:

`y - f(x_0) = f'(x_0)(x - x_0)`

The problem provides the coordinates of tangency point, hence you only need to find `f'(x_0)` such that:

`f'(x) = (6x+15-3e^x)' =gt f'(x) = 6 - 3e^x`

You need to substitute 0 for x in f'(x) such that:

`f'(0) = 6 - 3e^0 =gt f'(0) = 6-3=3`

You need to write the equation of tangent line such that:

`y - 12 = 3(x-0) `

You need to write the slope intercept form of the equation of tangent line, hence you need to isolate y to the left side such that:

`y = 3x + 12`

**Hence, evaluating the equation of tangent line yields`y =3x + 12` , thus `m=3` and `b=12` .**