Study of x³−4x and tangent lines at A and B

Text

Given the function $y = x^3 - 4x$:

• (a) find the intersections with the $x$-axis, study the sign and sketch an approximate graph;
• (b) let $A$ be the point of abscissa $2$ and $B$ the point of abscissa $0$; find the tangent lines to the graph at these points.

Solution

(a) Zeros and sign. Factoring: $y = x^3 - 4x = x(x-2)(x+2).$ The zeros are $x = -2,\ 0,\ 2$. Studying the sign of the product of the three factors: $y > 0 \ \text{on}\ (-2,0)\cup(2,+\infty), \qquad y < 0 \ \text{on}\ (-\infty,-2)\cup(0,2).$ The graph is a cubic increasing at its ends, crossing the $x$-axis at the three zeros.

(b) Tangent lines. The derivative is $y' = 3x^2 - 4$.

• At $A(2,\,0)$: $y'(2) = 3\cdot 4 - 4 = 8$, so the tangent is $y - 0 = 8(x-2) \ \Rightarrow\ y = 8x - 16.$
• At $B(0,\,0)$: $y'(0) = -4$, so the tangent is $y - 0 = -4(x-0) \ \Rightarrow\ y = -4x.$ $A$ and $B$ lie on the graph, so these are exactly the tangents at those points. $\boxed{\text{tangent at } A:\ y = 8x - 16;\qquad \text{tangent at } B:\ y = -4x}$