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 spot when a real situation needs a quadratic model. The usual cases are projectile motion (a thrown or fired object), maximum or minimum problems, and area-perimeter problems. Then you find the vertex (the turning point), the axis-intercepts, and the zeros, and say what each one means in the real situation. The deeper skill is reading the parabola as a story. It starts at the $y$-intercept, turns at the vertex, and ends at the zeros, and each point carries the units of the situation.

The answer

Figure 1. A downward-opening parabola $y = -(x+1)(x-5)$ illustrating the four features every quadratic question asks you to find: the vertex (maximum at $(2, 9)$), the axis of symmetry ($x = 2$), the two zeros ($x = -1$ and $x = 5$) and the $y$-intercept ($(0, 5)$).

Standard form

A quadratic has the form

The graph is always a parabola: a single smooth U-shaped (or upside-down U) curve with one turning point and a line of symmetry. Each of the three coefficients ($a$, $b$ and $c$, the numbers in the model) controls one feature, which is the key to reading a model quickly:

• $a$ controls the opening. $a > 0$ opens upward, so the vertex is a minimum; $a < 0$ opens downward, so the vertex is a maximum. The larger $|a|$ is, the narrower the parabola.
• $c$ is the $y$-intercept, the value when $x = 0$.
• $a$ and $b$ together fix the turning point through $x = -\frac{b}{2a}$.

The single most useful habit is to read the sign of $a$ first. In a profit, height or area problem the question almost always wants the biggest value, and a downward parabola ($a < 0$) is the only shape that has one.

Finding the vertex

The vertex (highest or lowest point) occurs at

Substitute this $x$ back into $y = a x^2 + b x + c$ to find the maximum or minimum value. Finding the $x$-coordinate is only half the job; the question usually wants the $y$, the actual maximum height or maximum profit.

The $x$-coordinate of the vertex is also the axis of symmetry, the vertical line $x = -\frac{b}{2a}$ that splits the parabola into two mirror-image halves. That symmetry is a free checking tool. Suppose two $x$-values give the same $y$, such as the two zeros, or two points you worked out either side. Then the axis of symmetry sits exactly halfway between them, so the vertex $x$ is just their average. For zeros at $x = -1$ and $x = 5$, the axis is at $x = \frac{-1 + 5}{2} = 2$ with no other working.

Zeros (where $y = 0$)

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

• By factorising if the quadratic factors cleanly. This is the level of factorising expected in HSC Maths Standard 2.
• By reading the graph. The zeros are where the parabola crosses the $x$-axis.
• From a table of values given in the question stem, look for rows where $y = 0$ or where $y$ changes sign (the zero lies between them).

Maths Standard 2 does not use the quadratic formula or the discriminant (a test for how many zeros there are). Those belong to Maths Advanced and Extension 1. If a model's quadratic does not factor cleanly, the question will give you the zeros on a graph or in a table, so you read them off rather than calculate them.

$y$-intercept

Set $x = 0$. The $y$-intercept is just $c$. In context it is the starting value: the height an object is launched from, the fixed cost before any sales, the area when a side length is zero.

Spotting a quadratic from a table: the constant second difference

A favourite exam task hands you a table and asks which model fits. The test for a quadratic is the second difference.

Take the projectile table $h = 0, 15, 20, 15, 0$ at $t = 0, 1, 2, 3, 4$. The first differences are each value minus the one before: $+15, +5, -5, -15$. These are not constant, so the model is not linear. Now take the differences of those differences, the second differences: $-10, -10, -10$. These are constant, so the model is quadratic. That constant value $-10$ equals $2a$, twice the leading coefficient (here $2 \times (-5)$), which is a neat check once you have the equation.

Sketch the parabola, stage by stage

The surest way to graph a quadratic model is to make a small table of values, plot the points, then join them with one smooth curve. The build below is for a ball thrown straight up whose height is $h = -5t^2 + 20t$ metres after $t$ seconds; the table gives $h = 0, 15, 20, 15, 0$ at $t = 0, 1, 2, 3, 4$.

Stage 1, draw and scale the axes. Put time $t$ on the horizontal axis (0 to 4 seconds, the flight time) and height $h$ on the vertical axis (0 to 20 metres, a little above the peak). Even spacing matters: equal steps must be equal distances or the curve will look lopsided.

Stage 2, plot the table points. Plot each pair from the table: $(0, 0)$, $(1, 15)$, $(2, 20)$, $(3, 15)$ and $(4, 0)$. Notice the up-then-down symmetry already showing in the dots, a tell-tale sign the model is quadratic.

Stage 3, join the points with one smooth curve. Draw a single smooth parabola through the points, not a series of straight line segments. The curve should be rounded at the top, not pointed, because a parabola turns gradually.

Stage 4, mark the vertex and axis of symmetry. The highest point is the vertex $(2, 20)$, the maximum height. The dashed vertical line $t = 2$ through it is the axis of symmetry: the curve is a mirror image either side of it, which is why $t = 1$ and $t = 3$ share the same height. The two zeros, where the curve meets the $t$ axis, are $t = 0$ (launch) and $t = 4$ (landing).

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 the initial vertical velocity, and $h_0$ is the initial height. The vertex gives the peak height and the time it is reached; the positive zero gives the landing time; the $h$-intercept $h_0$ is the launch height.
• Maximum revenue or profit. $R = p(x) \cdot x$ where $p(x)$ is a linear price-quantity model. Multiplied out, $R$ is quadratic, and because more sales usually mean a lower price, the parabola opens downward. The vertex gives the revenue-maximising quantity.
• Maximum area with fixed perimeter. A length-width problem with a constraint. Substitute the constraint into the area formula to get a quadratic in one variable; the vertex gives the dimensions of greatest area (a square or, against a wall, a half-square).

How exam questions ask about quadratic models

The wording varies but each phrasing maps to one feature of the parabola. Learn the translation:

• "Find the maximum height / maximum profit / greatest area." Find the vertex with $x = -\frac{b}{2a}$, then substitute back for the $y$. State that $a < 0$ so the vertex is a maximum.
• "How long until it hits the ground / when does it return to the start / for what price is there no profit?" These are the zeros. Factorise, or read them from the supplied graph or table, and keep only the physically sensible (usually positive) one.
• "What is the initial height / fixed cost / starting value?" That is the $y$-intercept, the constant $c$, the value at $x = 0$.
• "At what time / quantity does the maximum occur?" That is just the $x$-coordinate of the vertex, $-\frac{b}{2a}$.
• "Which model fits this table?" Check the differences: a constant second difference means quadratic.
• "Sketch the graph." Plot the intercepts and the vertex, then join with a smooth curve, and label all three features.
• "For how long is the height above $H$ metres?" Read the two $t$-values where the curve crosses the horizontal line $h = H$ off the graph and subtract; the answer is the width of the parabola at that height.