The point at which the tangent to the curve y | vedclass

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(B) Given the curve equation is $y = 2x^2 - x + 1$.To find the slope of the tangent,we differentiate the equation with respect to $x$:$\frac{dy}{dx} =

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$(1, 2)$

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(B) Given the curve equation is $y = 2x^2 - x + 1$.To find the slope of the tangent,we differentiate the equation with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(2x^2 - x + 1) = 4x - 1$.The given line is $y = 3x + 9$,which is in the form $y = mx + c$,where the slope $m = 3$.Since the tangent is parallel to the line $y = 3x + 9$,their slopes must be equal.Therefore,$4x - 1 = 3$.Solving for $x$:$4x = 4 \implies x = 1$.Now,substitute $x = 1$ into the original curve equation to find the $y$-coordinate:$y = 2(1)^2 - (1) + 1 = 2 - 1 + 1 = 2$.Thus,the required point is $(1, 2)$.

The point at which the tangent to the curve $y = 2x^2 - x + 1$ is parallel to $y = 3x + 9$ will be

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