Convex and concave curves and points of inflectionAQA A-Level Maths: Revision notes

The second derivative $\frac{d^2y}{dx^2}$ is the rate of change of the gradient, so its sign describes the shape of the curve.

• $\frac{d^2y}{dx^2}>0$: the gradient is increasing and the curve is convex (it curves upwards, like the bottom of a bowl).
• $\frac{d^2y}{dx^2}<0$: the gradient is decreasing and the curve is concave (it curves downwards, like the top of a hill). For $y=x^3-6x^2+5x$, $\frac{d^2y}{dx^2}=6x-12$. It is negative for $x<2$ (concave) and positive for $x>2$ (convex).

A point of inflection is a point where the curve changes from convex to concave or from concave to convex. At a point of inflection $\frac{d^2y}{dx^2}=0$ and $\frac{d^2y}{dx^2}$ changes sign. To find one: solve $\frac{d^2y}{dx^2}=0$, check the sign of $\frac{d^2y}{dx^2}$ either side, then find $y$ from the original equation. For $y=x^3-6x^2+5x$, $\frac{d^2y}{dx^2}=0$ at $x=2$, where it changes from negative to positive, and $y=8-24+10=-6$. The point of inflection is $(2,-6)$.

A curve can have more than one. For $y=x^4-4x^3+6$, $\frac{d^2y}{dx^2}=12x^2-24x=12x(x-2)$. This is zero at $x=0$ and $x=2$ and changes sign at both, so the points of inflection are $(0,6)$ and $(2,-10)$. The curve is convex for $x<0$ and $x>2$, and concave for $0<x<2$. For an expression with $x$ in a denominator, such as $y=x^2+\frac{8}{x}$, $\frac{d^2y}{dx^2}=2+\frac{16}{x^3}$; note the excluded value $x=0$ when you split the number line into regions.

A point of inflection may be stationary ($\frac{dy}{dx}=0$) or not. For $y=x^3$, the point of inflection at the origin is stationary. For $y=x^3-6x^2+5x$ the gradient at $(2,-6)$ is $3(4)-24+5=-7$, so it is not. A point of inflection is where the gradient is at its greatest or least value: it is least at the inflection of $y=x^3-6x^2+5x$, because the gradient decreases for $x<2$ and increases for $x>2$.

Use the condition $\frac{d^2y}{dx^2}=0$ at the given $x$ to find an unknown constant. For $y=x^3+ax^2+bx$ with an inflection at $x=1$: $6+2a=0$ gives $a=-3$. If the gradient there is $-5$ then $3-6+b=-5$, so $b=-2$, and the point is $(1,-4)$. For a tangent at the inflection, use $y-y_1=m(x-x_1)$ with $m=\frac{dy}{dx}$ at that point.