# Consider the function y = x^3 + 2x^2 - 6x + 5 ,the coordinates of the points on the graph where the tangent line is parallel to the line y = 2x + 8

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### 1 Answer

You should remember that the tangent line to the given curve, at the point you need to find, is derivative of the function `y = x^3 + 2x^2 - 6x + 5` at this point such that:

`(dy)/(dx)|_(x=x_0) = (3x^2 + 4x - 6)|_(x=x_0)`

Notice that the problem provides the information that the tangent line is parallel to the line `y = 2x + 8` , hence, the slopes of the tangent line and the given line needs to be equal such that:

`m_t = 2`

You need to remember that the slope of the tangent line has the same value to the slope of the given line such that:

`3x_0^2 + 4x_0 - 6 = 2 => 3x_0^2 + 4x_0 -8 =0`

You should use quadratic formula to find `x_0` such that:

`x_0 = (-4+-sqrt(16 + 96))/6 => x_0 = (-4+-sqrt(112))/6`

`x_0 = (-4+-4sqrt7)/6 => x_0 = (-2+-2sqrt7)/3`

You may evaluate y coordinate at `x_0 = (-2+-2sqrt7)/3` , substituting the values of `x_0` in equation of the curve such that:

`y_0 =(-2+2sqrt7)^3/27 + 2(-2+2sqrt7)^2/9 - 2(-2+2sqrt7) + 5`

`y_0 = -8 + 24sqrt7 - 168 + 56sqrt7 +24 - 24sqrt7 + 84+108 - 108sqrt7 + 135`

`y_0 = 175 - 52 sqrt7`

`y_0 = (-2-2sqrt7)^3/27 + 2(-2-2sqrt7)^2/9 - 2(-2-2sqrt7) + 5`

`y_0 = -8- 24sqrt7 - 168- 56sqrt7+ 24 + 24sqrt7 + 84- 108 +108sqrt7 + 135`

`y_0 =-41 + 52sqrt7`

**Hence, evaluating the coordinates of the points where the tangent line to the curve is parallel to the given line yields `(-2+2sqrt7,175 - 52 sqrt7), (-2+2sqrt7,-41 + 52sqrt7).` **