The slope of the tangent and normal to $y = x^{2} - 3 x + 5$ at $(2, 3)$ are __ and _ respectively.

1, 1

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1, - 1

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- 1, 1

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2, - 2

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Solution

The correct option is C $- 1, 1$
the slope of the tangent at any point is the derivative of the equation at that point

therefore

$\frac{d}{d x} (x^{2} - 3 x + 5) = 2 x - 3$

at $(2, 3)$ $\frac{d y}{d x}$ is $2 (2) - 3 = 1$

therefore the slope of the tangent is $1$

normal is perpendicular to the tangent

therefore the slope of the normal*slope of the tangent= $- 1$

therefore the slope of the normal is $- 1$

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