Find the point at which a given line is tangent to a given curve

Consider the curve given by the cubic equation

$y = x^3 - 6x^2 + 8x.$

Show that the line $y = -x$ is tangent to this curve and find the point of tangency. Determine if this line intersects the curve anywhere else.

First, we compute the derivative of the curve,

$y' = 3x^2 - 12x + 8.$

For the line $y = -x$ to be tangent to the curve $y = x^3 - 6x^2 + 8x$ it must be at a point $x$ such that $y' = -1$ (since this line has slope $-1$ , the derivative of the curve must be $-1$ ). So,

$y' = -1 \quad \implies \quad 3x^2 - 12x + 8 = -1 \quad \implies \quad x = \{ 1, 3\}.$

Next, for the line to be tangent to the curve the point must actually be on the curve. So, we test out the two possible values of $x$ .

If $x = 1$ , we have $y(1) = 1 - 6 + 8 = 3 \neq -1$ . So $y = -x$ is not tangent to the curve at $x = 1$ .

If $x = 3$ , we have $y(3) = 27 - 9(6) + 24 = -3 = -x$ ; hence, $y = -x$ is tangent to the curve at $(3,-3)$ .

This tangent line also intersects the curve at $(0,0)$ .

Stumbling Robot

Find the point at which a given line is tangent to a given curve

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