The tangent to the curve y = 2x^2 - x + 1 at | vedclass

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Question

(B) Given the curve equation is $y = 2x^2 - x + 1$.The slope of the tangent at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 4x - 1$.

Vedclass · Accepted Answer

$(1, 2)$

Vedclass · Answer

(B) Given the curve equation is $y = 2x^2 - x + 1$.The slope of the tangent at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 4x - 1$.The given line is $y = 3x + 4$,which is in the slope-intercept form $y = mx + c$,where the slope $m = 3$.Since the tangent is parallel to the line,their slopes must be equal. Therefore,we set the derivative equal to the slope of the line:$4x - 1 = 3$$4x = 4$$x = 1$Now,substitute $x = 1$ into the curve equation to find the $y$-coordinate of $P$:$y = 2(1)^2 - (1) + 1$$y = 2 - 1 + 1 = 2$Thus,the coordinates of point $P$ are $(1, 2)$.

The tangent to the curve $y = 2x^2 - x + 1$ at a point $P$ is parallel to the line $y = 3x + 4$. The coordinates of $P$ are:

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