Model practical situations with quadratic functions and find maximum or minimum values, intercepts and zeros

What this dot point is asking

NESA wants you to recognise when a real situation needs a quadratic model (projectile motion, maximum or minimum problems, area-perimeter problems), find the vertex, the axis-intercepts and the zeros, and interpret what each means in context.

The answer

Standard form

A quadratic has the form

The graph is a parabola.

• IMATH_5 : opens upward, vertex is a minimum.
• IMATH_6 : opens downward, vertex is a maximum.

Finding the vertex

The vertex (highest or lowest point) occurs at

Substitute back into $y = a x^2 + b x + c$ to find the maximum or minimum value.

The $x$-coordinate of the vertex is also the axis of symmetry; the parabola is mirror-symmetric about $x = -\frac{b}{2a}$.

Zeros (where $y = 0$)

Solve $a x^2 + b x + c = 0$.

• By factorising if the quadratic factors cleanly.
• By the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$ in general.

The discriminant $\Delta = b^2 - 4 a c$ tells you how many real zeros there are: two if $\Delta > 0$, one if $\Delta = 0$, none if $\Delta < 0$.

IMATH_17 -intercept

Set $x = 0$. The $y$-intercept is just $c$.

Practical contexts

• Projectile motion. $h = -\frac{g}{2} t^2 + v_0 t + h_0$ where $g \approx 9.8$ or $10$ m/s$^2$, $v_0$ is initial vertical velocity, $h_0$ is initial height. Vertex gives the peak height; positive zero gives the landing time.
• Maximum revenue or profit. $R = p(x) \cdot x$ where $p(x)$ is a linear price-quantity model; multiplied out, $R$ is quadratic. Vertex gives the revenue-maximising quantity.
• Maximum area with fixed perimeter. Length-width problems with a constraint. Substitute the constraint into the area formula to get a quadratic.