The quadratic function

Throw a ball in the air. It goes up, slows down, stops for an instant and falls back; the path is a curve called a parabola. The same curve appears in bridge arches, satellite dishes and the graph of profit against price. All these situations are described by a quadratic function, and knowing its highest or lowest point answers the question "what is the best we can do?".

The function $y=ax^2+bx+c$

The number $|a|$ controls how narrow the parabola is.

Sketching a parabola

To sketch $y=ax^2+bx+c$: (1) find the vertex; (2) draw the axis; (3) find the points where the graph meets the axes ($x=0$ gives $y=c$; $y=0$ gives the roots, if any); (4) use symmetry about the axis.

Table of variation

The table of variation shows where the function goes up or down. For $y=x^2-4x+3$ ($a>0$):

$x$	$-\infty$		$2$		$+\infty$
$y$	$+\infty$	decreasing	$-1$	increasing	$+\infty$

For $a>0$: decreasing on $\left(-\infty, -\frac{b}{2a}\right)$, increasing on $\left(-\frac{b}{2a}, +\infty\right)$, minimum $-\frac{\Delta}{4a}$. For $a<0$ it is the other way round and $-\frac{\Delta}{4a}$ is the maximum.

Finding the formula

Translations: $y=a(x-p)^2+q$

Completing the square gives $ax^2+bx+c=a(x-p)^2+q$ with $p=-\frac{b}{2a}$ and $q=-\frac{\Delta}{4a}$. So the vertex is $(p, q)$. The graph is the parabola $y=ax^2$ moved $p$ units to the right and $q$ units up (left or down if negative).

Expanding, $2(x-1)^2+3=2x^2-4x+5$, which has vertex $\left(\frac{4}{4}, \frac{-(16-40)}{8}\right)=(1, 3)$. ✓

Applications

Key terms

Summary

• $y=ax^2+bx+c$ has a parabola as its graph, with vertex $\left(-\frac{b}{2a}, -\frac{\Delta}{4a}\right)$ and axis $x=-\frac{b}{2a}$.
• $a>0$: opens up, minimum at the vertex. $a<0$: opens down, maximum at the vertex.
• Sketch with the vertex, the intercepts and symmetry; summarise with a table of variation.
• $y=a(x-p)^2+q$ has vertex $(p, q)$.
• Use the vertex to solve maximum and minimum problems; always check the allowed values of $x$.

Practice

Show solutions

1. $x_I=3$, $y_I=9-18+5=-4$. Vertex $(3, -4)$, axis $x=3$. $x^2-6x+5=0$ gives $x=1$ or $x=5$.
2. $a=-2<0$: opens down. $x_I=-\dfrac{4}{-4}=1$, $y_I=-2+4+1=3$. Vertex $(1, 3)$.
3. $a=3>0$, so there is a minimum at $x=\dfrac{12}{6}=2$: $y=12-24+7=-5$. Minimum $-5$ at $x=2$.
4. $a=-1<0$, vertex at $x=2$, $y=-4+8=4$.

$x$	$-\infty$		$2$		$+\infty$
$y$	$-\infty$	increasing	$4$	decreasing	$-\infty$

Maximum $4$ at $x=2$. 5. $y=a(x-1)^2-4$. Through $(3, 0)$: $4a-4=0$, $a=1$. So $y=(x-1)^2-4=x^2-2x-3$. 6. $c=1$ (from $A$). $B$: $a+b+1=0$. $C$: $4a+2b+1=3$, so $2a+b=1$. Subtract: $a=2$, $b=-3$. $y=2x^2-3x+1$. Check $C$: $8-6+1=3$ ✓. 7. $x^2-2x-3=(x-1)^2-4$. Move the parabola $y=x^2$ one unit to the right and four units down. 8. Maximum at $t=-\dfrac{10}{-10}=1$ s: $h(1)=-5+10=5$ m. $h=0$: $-5t(t-2)=0$, so it lands at $t=2$ s. 9. Let the sides be $x$ and $30-x$. Area $S=x(30-x)=-x^2+30x$, maximum at $x=15$, $S=225$. It is a square of side $15$ m, area $225$ m$^2$. 10. $y=(x-m)^2+3-m^2$, so the minimum is $3-m^2$. Setting $3-m^2=2$ gives $m^2=1$, so $m=1$ or $m=-1$.