At what points on the curve y = 2x2 – x + 1 is the tangent parallel

1.	At what points on the curve y = 2x2 – x + 1 is the tangent parallel to the line y = 3x + 4?
Answer» 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 ⇒ \(\frac{dy}{dx}\) = \(\frac{d}{dx}\)(2x²) – \(\frac{d}{dx}\)(x) + \(\frac{d}{dx}\)(1) ⇒ \(\frac{dy}{dx}\) = 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)

At what points on the curve y = 2x2 – x + 1 is the tangent parallel to the line y = 3x + 4?

Answer»

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

⇒ \(\frac{dy}{dx}\) = \(\frac{d}{dx}\)(2x²) – \(\frac{d}{dx}\)(x) + \(\frac{d}{dx}\)(1)

⇒ \(\frac{dy}{dx}\) = 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)