Cubic graphs

A stationary point is where $f^{\prime }(x)=0$. On a cubic this is usually a local maximum or minimum (a turning point).

To find the turning points of $f(x)=x^3-3x^2-9x+5$:

$f^{\prime }(x)=3x^2-6x-9=3(x-3)(x+1)=0$

$x=3\text{ or }x=-1$

Substitute back into $f$: $f(-1)=10$ and $f(3)=-22$. The turning points are $(-1;10)$ and $(3;-22)$.

To decide which is a maximum, use the second derivative: $f^{\prime \prime }(x)=6x-6$. $f^{\prime \prime }(-1)=-12<0$ gives a local maximum, and $f^{\prime \prime }(3)=12>0$ gives a local minimum.

The point of inflection is where the concavity changes. For a cubic it is where $f^{\prime \prime }(x)=0$, and it lies exactly halfway between the two turning points.

• $f^{\prime \prime }(x)>0$: concave up (the graph bends upwards like a cup)
• $f^{\prime \prime }(x)<0$: concave down

Example. For $f(x)=x^3-3x^2-9x+5$, $f^{\prime \prime }(x)=6x-6=0$ gives $x=1$, and $f(1)=-6$. The point of inflection is $(1;-6)$. The graph is concave up for $x>1$ and concave down for $x<1$.

A cubic such as $y=x^3$ has a stationary point of inflection: $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$, but the graph does not turn.

For a full sketch, show:

1. Shape: $a>0$ rises to the right, $a<0$ falls to the right.
2. $y$-intercept: let $x=0$.
3. $x$-intercepts: factorise $f(x)=0$. Use the factor theorem to find the first factor: test values that divide the constant term.
4. Turning points: solve $f^{\prime }(x)=0$.
5. Point of inflection, if asked.

Factor theorem example. $f(x)=x^3-7x-6$. $f(-1)=-1+7-6=0$, so $(x+1)$ is a factor. Dividing gives $f(x)=(x+1)(x^2-x-6)=(x+1)(x-3)(x+2)$. The factorising cubics topic covers the remainder and factor theorems in full.

Label every intercept and turning point with its coordinates. Unlabelled sketches lose marks even when the shape is right.

Use the $x$-intercepts to write $f(x)=a(x-x_1)(x-x_2)(x-x_3)$, then use one more point to find $a$.

If the graph touches the $x$-axis at $x=p$ (a turning point on the axis), that root is repeated: $f(x)=a(x-p{)}^2(x-q)$.

Example. A cubic cuts the $x$-axis at $-1$, $2$ and $4$ and the $y$-axis at 16.

$f(x)=a(x+1)(x-2)(x-4)$

$f(0)=a(1)(-2)(-4)=8a=16\Rightarrow a=2$

If you are given turning points instead, use $f^{\prime }(x)=0$ at those $x$-values to form equations in the unknown coefficients.

The same transformation rules apply as for other functions:

• $y=f(x-p)$ shifts the graph $p$ units right.
• $y=f(x)+q$ shifts the graph $q$ units up.
• $y=-f(x)$ reflects in the $x$-axis, and $y=f(-x)$ reflects in the $y$-axis.

Every point moves, including the turning points. If $f$ has a turning point at $(-1;10)$, then $y=f(x-2)+3$ has a turning point at $(1;13)$.

A common question asks for values of $k$ so that $f(x)=k$ has a certain number of roots. Draw horizontal lines $y=k$ across the graph: three roots between the turning point $y$-values, two roots at those values, and one root outside them.