The tangent to the curve y 2x2 ndash x 1 is parallel to the line y 3x 9 at the point ........

The tangent to the curve y = 2x²–x+1 is parallel to the line y =3x + 9 at the point

3,9)

(2,-1)

(2,1)

(1,2)

Let’s break it down:

- First, the line y = 3x + 9 has a slope of 3. So, our tangent needs to match that slope.

- The curve y = 2x² – x + 1 has a derivative (slope) of dy/dx = 4x – 1.

- Set that equal to 3:

4x – 1 = 3 ? 4x = 4 ? x = 1

- Plug x = 1 back into the curve:

y = 2(1)² – 1 + 1 = 2 – 1 + 1 = 2

- So, the tangent is parallel to the line at (1, 2).

Options:

- (1, 3, 9): Not a point, just numbers.

- (2, –1): x = 2, slope is 4(2) – 1 = 7, which isn’t 3.

- (2, 1): Same logic, wrong y-value.

- (1, 2): Correct.

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