# Find a unit vector that is parallel to the tangent line to the parabola y=-2x-x^2 at the point...

## Question:

Find a unit vector that is parallel to the tangent line to the parabola {eq}y=-2x-x^2 {/eq} at the point (1,-3).

## Parallel and Perpendicular vectors:

In vector calculus, slope is used to find the vector parallel to a tangent line to the curve. Slope of a curve {eq}y=f(x) {/eq} is {eq}m=f'(x) {/eq}. Now using the slope, parallel vector is expressed by the formula {eq}\mathbf i+m \mathbf j {/eq} and perpendicular vector is calculated by {eq}m\mathbf i-\mathbf j {/eq}

## Answer and Explanation:

{eq}y=-2x-x^2 {/eq}

Slope {eq}\frac{dy}{dx}= -2-2x {/eq}

At {eq}(1,-3) {/eq}

{eq}\frac{dy}{dx}= -4 {/eq}

Vector parallel to tangent line...

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