Write the coordinates of the point at which the tangent to the curve `y=2x^2-x+1` is parallel to the line `y=3x+9` .

Explanation

We have `y = 2x^2 - x + 1`

⇒ `dy/dx = 4x - 1 `

We know that this equation gives the slope of tangent to the curve.

Since this tangent is parallel to y = 3x + 9,

the slope of the tangent is 3 and so

4x - 1= 3 or x =1.

Therefore `y = 2x^2 - x + 1 = 2.` Thus, the point (x, y) is (1, 2)