At what points on

Given:

The curve is y = 2x² – x + 1and the line y = 3x + 4

First, we will find The Slope of tangent

y = 2x² – x + 1

⇒ (2x²) – (x) + (1)

⇒ = 4x – 1 ...(1)

y = 3x + 4 is the form of equation of a straight line y = mx + c,where m is the The Slope of the line.

so the The Slope of the line is y = 3(x) + 4

Thus, The Slope = 3. ...(2)

From (1) & (2),we get,

4x – 1 = 3

⇒ 4x = 4

⇒ x = 1

Substituting x = 1in y = 2x² – x + 1,we get,

⇒ y = 2(1)² – (1) + 1

⇒ y = 2 – 1 + 1

⇒ y = 2

Thus, the required point is (1,2)