Verified answer

Answer:

(b) (1/2)

Step-by-step explanation:

To find the point at which the tangent to the curve y=2x^{2}-x+1 is parallel to 3x+9, we need to find the value of x at which the derivative of y=2x^{2}-x+1 is equal to the slope of 3x+9, which is 3.                            (3x+9)' = 3

The derivative of y=2x^{2}-x+1 is 4x-1.                                [2(2)x-1]

So, we need to solve the equation 4x-1=3 for x.

4x-1=3

4x=4

x=1

Therefore, the point at which the tangent to the curve y=2x^{2}-x+1 is parallel to 3x+9 is when x=1.

To find the corresponding y-value, we can substitute x=1 into the equation y=2x^{2}-x+1,

which gives,

y=2(1)^{2}-(1)+1

y=2.

So, the point is (1,2).