The tangent to the curve $y = 2x^{2} - x + 1$ at a point $P$ is parallel to $y = 3x + 4$, then the coordinates of $P$are

Question

Dr. Abubacker · Accepted Answer

(b)
Given curve is $y = 2x^{2} - x + 1$
Let the coordinate of $P$ are $(h,\ k)$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 4x - 1$
At the point $(h,\ k)$, the slope
$= \left( \frac{dy}{dx} ight)_{(h,\ k)} = 4h - 1$
Since, the tangent is parallel to the given line $y = 3x + 4$
$\Rightarrow \ \ 4h - 1 = 3\ \  \Rightarrow \ \ h = 1,\ k = 2$
$	herefore$ Coordinates of point $P$ are (1, 2)

Q. The tangent to the curve $y = 2x^{2} - x + 1$ at a point $P$ is parallel to $y = 3x + 4$, then the coordinates of $P$are

Explanation

Reviewed by Dr. Abubacker

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