Question

How do you find the slope of a tangent line to the graph of the function `4x^2 + y^2 = 4` at (0, -2)?

Answer 1

Start by differentiating the relation with respect to x.

Explanation:

You must use implicit differentiation for this one:

`d/dx(4x^2 + y^2) = d/dx(4)`

`d/dx(4x^2) + d/dx(y^2) = d/dx(4)`

`8x + 2y(dy/dx) = 0`

`2y(dy/dx) = -8x`

`dy/dx = -(4x)/y`

The slope of the tangent is given by evaluating `f'(a)`, where `x = a` is the given point and `f'(x)` is the derivative of the function/relation.

Since this relation involves both x and y, we must plug in both the x and the y points to determine the slope of the tangent.

`m_"tangent" = -(4(0))/-2`

`m_"tangent" = 0`

Hence, the slope of the tangent is `0`. If you would like to go a step further, this means that the tangent is horizontal at the point `(0, -2)`, because a slope of `0` denotes a horizontal line.

Hopefully this helps!