Question

The tangent to the curve y = 2x² - x + 1 at point P is parallel to y = 3x + 4, the coordinates of point P are

(A) (2, 1)

(B) (1, 2)

(C) (− 1, 2)

(D) (2, − 1)

Answer 1

Answer is (B) (1, 2)

We have

y = 2x² - x + 1

Let the coordinates of P be (h, k). Then

(dy/dx)_{(h, k)} = 4h - 1

Clearly, P is parallel to y = 3x + 4.

Therefore, slopes are equal

4h - 1 = 3 ⇒ h = 1

Therefore, P is (1, 2).