# Thread: Equation of a tangent line

1. ## Equation of a tangent line

I have the following exponential equation
(x^3 + 2)(x^2 + X)

Find the equation of the tangent line when x=1

This is what I do to get the derivative

(3x^2)(x^2 + x) + (2x + 1)(x^3 + 2)

When I factor this out I get an answer that includes x^4 and x^3 and x^2 yet the answer for the equation of the line is 15x-9.

Where am I going wrong? I am really lost on tangent line equations.

2. Originally Posted by CMH
I have the following exponential equation
(x^3 + 2)(x^2 + X)

Find the equation of the tangent line when x=1

This is what I do to get the derivative

(3x^2)(x^2 + x) + (2x + 1)(x^3 + 2)

When I factor this out I get an answer that includes x^4 and x^3 and x^2 yet the answer for the equation of the line is 15x-9.

Where am I going wrong? I am really lost on tangent line equations.
The derivative, at x= 1, will give you the slope of the tangent.

calculate dy/dx and put x=1 in it to get the slope of tangent.

$slope, \;m = \left[\frac{dy}{dx}\right]_{x=1} = [3(1)^2][(1)^2 + 1] + [2(1) + 1][(1)^3 + 2]=15$

After that calculate the y - coordinate of the point, by putting x=1 in the eqn of curve, you will get (1, 6)

$y= [(1)^3 + 2][(1)^2 + 1]= 6$

Now, find the eqn of tangent using point slope form of line,

$y-y_1=m(x-x_1)$

y - 6 = 15 (x - 1)

y = 15x - 9

3. ## Thank you

Now it makes sense. Thank you so much.