What are the three discriminant cases?

Each parabola opens upward, but they sit at different heights relative to the x-axis. With > 0 the curve dips below the axis and crosses it twice; with = 0 the vertex sits exactly on the axis, giving one repeated root; with < 0 the whole curve stays above the axis, so there are no real roots.

NSWMaths AdvancedSyllabus dot point

How do you sketch a parabola from its equation, find its turning point and axis of symmetry, decide how many times it crosses the x-axis, and read off its maximum or minimum value?

Sketch a parabola by finding its intercepts by factoring, complete the square to write a quadratic in vertex form and read off the turning point and axis of symmetry, use the quadratic formula and the discriminant to find and classify the roots, and find the maximum or minimum value of a quadratic

A Year 11 Maths Advanced answer on the parabola: sketch by factoring for the intercepts, complete the square for the vertex form and turning point, use the quadratic formula and the discriminant to count and classify the roots, find the axis of symmetry, and read off the maximum or minimum value, with worked examples and practice questions.

Generated by Claude Opus 4.824 min answerUpdated 2026-06-21

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Quick answer

A quadratic $y = ax^2 + bx + c$ graphs as a parabola: concave up if $a > 0$ (a minimum) and concave down if $a < 0$ (a maximum), with $y$ -intercept $c$ . To sketch, factor and set $y = 0$ for the $x$ -intercepts, take the axis of symmetry halfway between them (or use $x = -\dfrac{b}{2a}$ ), and substitute to get the vertex. Completing the square rewrites the quadratic as $y = a(x - h)^2 + k$ , where the vertex is $(h, k)$ and $k$ is the minimum or maximum value; for a non-monic quadratic, factor $a$ out of the $x^2$ and $x$ terms first. The quadratic formula $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$ solves any quadratic, and the discriminant $\Delta = b^2 - 4ac$ counts the real roots: $\Delta > 0$ gives two (cuts the axis twice), $\Delta = 0$ gives one repeated root (touches the axis), and $\Delta < 0$ gives none (misses the axis); a perfect-square $\Delta$ means rational roots, otherwise surds. In real contexts (projectile height, the area enclosed by a fixed length of fence) the maximum or minimum value is the vertex, found from $x = -\dfrac{b}{2a}$ , with any negative time or length rejected.

What this dot point is asking

A quadratic function is any function you can write as $y = ax^2 + bx + c$ with $a \ne 0$ , and its graph is always a parabola: a single smooth U-shaped curve. This dot point is the complete toolkit for that curve. You need to sketch it (the intercepts, the turning point and the way it opens), to locate its vertex exactly using completing the square, to decide how many times it cuts the $x$ -axis using the discriminant, and to read off its maximum or minimum value. These are not four unrelated skills: they are four views of the same curve, and the strongest answers move between them fluently.

The reason the parabola matters so much in the HSC is that it is the first genuinely curved graph you can analyse completely by hand, and the methods here come back everywhere. Projectile motion, the area you can enclose with a fixed length of fence, the revenue a business earns as it changes its price: all are quadratics, and all are solved by finding the vertex. The marks in an exam live in choosing the right tool for the wording (factor to get intercepts; complete the square to get the vertex; use the discriminant to count roots) and in laying the working out cleanly. This page builds each tool and then shows exactly which question wordings call for which.

The answer

The parabola: shape, concavity and the basic features

Every quadratic graph is a parabola, and three features fix it. The concavity is set by the sign of $a$ : if $a > 0$ the parabola is concave up (opens upward, like a valley, with a lowest point), and if $a < 0$ it is concave down (opens downward, like a hill, with a highest point). The size of $a$ controls how narrow or wide the curve is, but not which way it opens. The $y$ -intercept is always $c$ , found by putting $x = 0$ . And the vertex (or turning point) is the single lowest point of a concave-up parabola or the single highest point of a concave-down one; the vertical line through it is the axis of symmetry, and the curve is a perfect mirror image in that line.

That symmetry is the single most useful fact about a parabola. It means the two $x$ -intercepts (when there are two) are the same distance either side of the axis, so the axis of symmetry sits exactly halfway between them. It also means that once you know the vertex and any one other point, you know a second point for free by reflecting across the axis. Every method below is really a different way of locating the vertex and the intercepts.

Sketching by factoring: the x-intercepts

The fastest sketch comes from the factored form. The $x$ -intercepts (also called the zeros or roots) are where the curve meets the $x$ -axis, so they are the values of $x$ that make $y = 0$ . If the quadratic factors, set each factor to zero. For $y = x^2 - 2x - 8$ , factor to $y = (x - 4)(x + 2)$ , so $y = 0$ when $x = 4$ or $x = -2$ ; the $x$ -intercepts are $(-2, 0)$ and $(4, 0)$ .

Once you have the two intercepts, the axis of symmetry is halfway between them, here at $x = \dfrac{-2 + 4}{2} = 1$ . Substituting that $x$ back into the equation gives the $y$ -coordinate of the vertex: $y = (1)^2 - 2(1) - 8 = -9$ , so the vertex is $(1, -9)$ . With the two intercepts, the $y$ -intercept $c = -8$ , and the vertex, you have more than enough to draw the curve. The diagram below builds that sketch up one stage at a time.

Stage 1, the framework: Draw the axes and plot the easiest point, the $y$ -intercept at $(0, -8)$ (just read off $c$ ). Because $a = 1 > 0$ , you already know the finished curve will be concave up.
Stage 2, the $x$ -intercepts: Factor $y = (x - 4)(x + 2)$ and set $y = 0$ to get the roots $x = -2$ and $x = 4$ . Plot $(-2, 0)$ and $(4, 0)$ on the $x$ -axis. These two points pin down where the curve crosses.
Stage 3, the vertex and axis of symmetry: The axis is halfway between the roots, $x = \dfrac{-2 + 4}{2} = 1$ ; substituting gives the vertex $(1, -9)$ . Draw the dashed axis of symmetry and plot the turning point.
Stage 4, the finished curve: Join the points with a smooth concave-up parabola, symmetric about $x = 1$ , passing through both intercepts, the $y$ -intercept and the vertex.

Completing the square: the vertex form

Factoring is fast but limited; completing the square works on every quadratic and hands you the vertex directly. The aim is to rewrite $y = ax^2 + bx + c$ in vertex form

y = a(x - h)^2 + k,

because in this form the vertex is simply $(h, k)$ and the axis of symmetry is $x = h$ . The reason is one fact about squares: $(x - h)^2$ is never negative and equals $0$ only when $x = h$ . So for a concave-up parabola ( $a > 0$ ), the smallest $y$ can be is $k$ , reached exactly at $x = h$ , which is the vertex; for a concave-down parabola ( $a < 0$ ), $k$ is the largest value instead.

For a monic quadratic (where $a = 1$ ) the procedure is: take the coefficient of $x$ , halve it, square it, then add and subtract that number. For $y = x^2 + 6x + 1$ , half of $6$ is $3$ and $3^2 = 9$ , so

y = (x^2 + 6x + 9) - 9 + 1 = (x + 3)^2 - 8.

The vertex is $(-3, -8)$ and the axis of symmetry is $x = -3$ . For a non-monic quadratic (where $a \ne 1$ ), first take the coefficient of $x^2$ out of the first two terms as a factor, complete the square inside, then tidy. For $y = 2x^2 - 12x + 7$ ,

y = 2(x^2 - 6x) + 7 = 2\big((x - 3)^2 - 9\big) + 7 = 2(x - 3)^2 - 11,

so the vertex is $(3, -11)$ . Completing the square also lets you find the $x$ -intercepts when no neat factors exist: set $y = 0$ and solve $a(x - h)^2 + k = 0$ by taking square roots, which naturally produces surd answers when they occur.

The axis of symmetry and the maximum or minimum value

You do not always have to complete the square to find the turning point. There is a direct formula for the axis of symmetry:

x = -\frac{b}{2a}.

This is just the $x$ -coordinate $h$ of the vertex; completing the square in general gives exactly this value. Once you have it, substitute it back into the quadratic to get the $y$ -coordinate of the vertex, which is the maximum or minimum value of the function. So for any quadratic the recipe for the turning point is two lines: compute $x = -\dfrac{b}{2a}$ , then substitute. For $y = 2x^2 - 12x + 7$ the axis is $x = -\dfrac{-12}{2 \times 2} = 3$ , and substituting gives $y = -11$ , matching the completed square above.

Whether that value is a maximum or a minimum is decided once more by the sign of $a$ : a concave-up parabola ( $a > 0$ ) has a minimum value at the vertex, and a concave-down parabola ( $a < 0$ ) has a maximum value. This is exactly what "find the maximum / minimum value" questions are asking for, and it is why a parabola is the natural model for any "greatest area" or "greatest height" problem.

The quadratic formula and the discriminant

Not every quadratic factors, so NESA expects the quadratic formula, which solves $ax^2 + bx + c = 0$ in every case:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

The quantity under the square root, $b^2 - 4ac$ , is so important it has its own name and symbol: the discriminant, written $\Delta$ (the Greek capital delta). It "discriminates" between the cases because its sign decides how many real roots exist (and therefore how many times the parabola crosses the $x$ -axis):

if $\Delta > 0$ there are two distinct real roots (the parabola cuts the $x$ -axis twice);
if $\Delta = 0$ there is exactly one real root, a repeated root (the parabola just touches the $x$ -axis at its vertex, so the $x$ -axis is a tangent);
if $\Delta < 0$ there are no real roots (the parabola never reaches the $x$ -axis), because a negative number has no real square root.

When $\Delta > 0$ and the coefficients are rational, you can discriminate one step further: if $\Delta$ is a perfect square the roots are rational (and the quadratic would have factored), and if it is not a perfect square the roots are surds. The good habit is to compute $\Delta$ first whenever a question asks "how many solutions" or "show that there are no solutions", because the count is the answer and you may not need the roots at all. The three cases are shown below.

The three discriminant cases. Each parabola opens upward, but they sit at different heights relative to the $x$ -axis. With $\Delta > 0$ the curve dips below the axis and crosses it twice; with $\Delta = 0$ the vertex sits exactly on the axis, giving one repeated root; with $\Delta < 0$ the whole curve stays above the axis, so there are no real roots.

Quadratics in real contexts

Quadratics are the natural model for anything that rises and then falls (or falls and then rises), and the vertex is almost always the point a worded question is hunting for. Projectile motion is the classic case: a stone thrown from a $15$ m cliff with height $h = -5t^2 + 10t + 15$ metres after $t$ seconds is a concave-down parabola, so its greatest height is the maximum value at the vertex. The axis is $t = -\dfrac{10}{2 \times (-5)} = 1$ second, and substituting gives $h = 20$ m, so the stone peaks at $20$ m after one second; setting $h = 0$ and solving $t^2 - 2t - 3 = 0$ gives $t = 3$ , the time it hits the sea. The coefficient $-5$ comes from gravity, the $+10t$ from the launch speed, and the $+15$ from the starting height, and each maps onto a feature of the curve.

Optimisation with a fixed resource is the other recurring context. If a farmer has $60$ m of fencing for a rectangular paddock with one side $x$ , the other side is $30 - x$ and the area is $A = x(30 - x) = -x^2 + 30x$ , a concave-down parabola. Its maximum value is at the axis $x = -\dfrac{30}{2 \times (-1)} = 15$ , giving the other side $30 - 15 = 15$ and the greatest area $A = 225 \text{ m}^2$ : a $15 \text{ m} \times 15 \text{ m}$ square. Reading a worded problem as "build the quadratic, then find its vertex" turns almost every "greatest" or "least" question into the same two-line method.

How exam questions ask about quadratics

The wording tells you which tool to reach for:

"Sketch the parabola / graph $y = \ldots$ , showing the intercepts and the vertex." Read off the concavity and $y$ -intercept, factor for the $x$ -intercepts, then the axis of symmetry (halfway between them, or $-\dfrac{b}{2a}$ ) and the vertex. Mark all of these on the curve, since each is a method mark.
"Write $y = \ldots$ in the form $y = (x - h)^2 + k$ " or "express in completed-square form." Complete the square; the vertex is then $(h, k)$ .
"Find the coordinates of the vertex / turning point" or "find the axis of symmetry." Use $x = -\dfrac{b}{2a}$ and substitute, or complete the square; either is acceptable unless the form is specified.
"Find the maximum / minimum value of ..." Find the vertex; the $y$ -coordinate is the value. State whether it is a maximum ( $a < 0$ ) or a minimum ( $a > 0$ ), and the $x$ at which it occurs if asked.
"How many solutions / real roots / $x$ -intercepts does ... have?" Compute the discriminant $\Delta = b^2 - 4ac$ and quote the case. You usually do not need to find the roots.
"Show that ... has no real roots" or "... has equal roots / a repeated root." Show $\Delta < 0$ for no roots, or $\Delta = 0$ for a repeated root.
"Find the value(s) of $k$ for which ... has equal roots / two roots / no roots." Form the discriminant in terms of $k$ , then set $\Delta = 0$ (equal), $\Delta > 0$ (two) or $\Delta < 0$ (none) and solve.
"Solve $ax^2 + bx + c = 0$ , giving the answer in exact form." Factor if it factors; otherwise use the quadratic formula and leave surds simplified.
"A ball is thrown ... find the greatest height" or "... find the maximum area." Build the quadratic from the context, then find the vertex; reject any physically impossible solution (such as a negative time or length).

Worked example

Six examples: sketching by factoring, completing the square for the vertex and range, completing the square on a non-monic quadratic, solving with the quadratic formula, counting roots with the discriminant, and a projectile maximum in context.

Sketch the parabola $y = x^2 - 2x - 8$ by factoring

Find the concavity, the intercepts, then the vertex.

Note the concavity and the $y$ -intercept. Here $a = 1 > 0$ , so the parabola is concave up. Putting $x = 0$ gives the $y$ -intercept $c = -8$ , the point $(0, -8)$ .

Factor for the $x$ -intercepts. Look for two numbers multiplying to $-8$ and adding to $-2$ : those are $-4$ and $2$ , so

y = (x - 4)(x + 2).

Setting $y = 0$ gives $x = 4$ or $x = -2$ , so the $x$ -intercepts are $(-2, 0)$ and $(4, 0)$ .

Find the vertex. The axis of symmetry is halfway between the roots, $x = \dfrac{-2 + 4}{2} = 1$ . Substituting $x = 1$ :

y = (1)^2 - 2(1) - 8 = 1 - 2 - 8 = -9,

so the vertex is $(1, -9)$ .

State the sketch. A concave-up parabola through $(-2, 0)$ and $(4, 0)$ , $y$ -intercept $(0, -8)$ , vertex $(1, -9)$ , symmetric about $x = 1$ . Its minimum value is $-9$ .

Complete the square for $y = x^2 + 6x + 1$ to find the vertex and range

Rewrite in vertex form, then read off the features.

Complete the square. The coefficient of $x$ is $6$ ; half of it is $3$ and $3^2 = 9$ , so add and subtract $9$ :

y = (x^2 + 6x + 9) - 9 + 1 = (x + 3)^2 - 8.

Read off the vertex and axis: In the form $y = (x - h)^2 + k$ the vertex is $(h, k) = (-3, -8)$ and the axis of symmetry is $x = -3$ .
State the minimum value and the range: Since $(x + 3)^2 \ge 0$ and is $0$ only at $x = -3$ , the smallest value of $y$ is $-8$ . So the minimum value is $-8$ , and the range is $y \ge -8$ .
Find the $x$ -intercepts in exact form: Put $y = 0$ : $(x + 3)^2 = 8$ , so $x + 3 = \pm\sqrt{8} = \pm 2\sqrt{2}$ , giving $x = -3 \pm 2\sqrt{2}$ . (These are surds, which is why the quadratic did not factor neatly.)

Complete the square for the non-monic $y = 2x^2 - 12x + 7$

When $a \ne 1$ , factor the leading coefficient out of the first two terms first.

Factor out the leading coefficient from the $x^2$ and $x$ terms.

y = 2(x^2 - 6x) + 7.

Complete the square inside the bracket. Half of $-6$ is $-3$ and $(-3)^2 = 9$ :

y = 2\big((x - 3)^2 - 9\big) + 7.

Expand the factor back through and tidy.

y = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11.

State the vertex and minimum. The vertex is $(3, -11)$ , the axis of symmetry is $x = 3$ , and since $a = 2 > 0$ the minimum value is $-11$ . As a check, the formula gives the axis directly as $x = -\dfrac{-12}{2 \times 2} = 3$ , matching.

Solve $2x^2 + 3x - 5 = 0$ with the quadratic formula

Always compute the discriminant first.

Identify the coefficients and the discriminant. Here $a = 2$ , $b = 3$ , $c = -5$ , so

\Delta = b^2 - 4ac = 3^2 - 4(2)(-5) = 9 + 40 = 49.

Since $\Delta = 49 > 0$ and is a perfect square, there are two rational roots.

Apply the formula.

x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm \sqrt{49}}{2 \times 2} = \frac{-3 \pm 7}{4}.

Compute both roots.

x = \frac{-3 + 7}{4} = \frac{4}{4} = 1 \qquad \text{or} \qquad x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}.

State the answer. The solutions are $x = 1$ and $x = -\dfrac{5}{2}$ . (As the discriminant was a perfect square, the quadratic also factors as $(x - 1)(2x + 5)$ , confirming both roots.)

Use the discriminant to count the roots of three quadratics

The sign of $\Delta$ alone answers a "how many roots" question.

(a) $y = 3x^2 - 5x + 1$ . With $a = 3$ , $b = -5$ , $c = 1$ ,

\Delta = (-5)^2 - 4(3)(1) = 25 - 12 = 13 > 0,

so there are two distinct real roots. Since $13$ is not a perfect square, they are surds.

(b) $y = 4x^2 - 12x + 9$ . With $a = 4$ , $b = -12$ , $c = 9$ ,

\Delta = (-12)^2 - 4(4)(9) = 144 - 144 = 0,

so there is exactly one repeated real root (the parabola touches the $x$ -axis). Indeed $4x^2 - 12x + 9 = (2x - 3)^2$ .

(c) $y = 2x^2 + x + 3$ . With $a = 2$ , $b = 1$ , $c = 3$ ,

\Delta = (1)^2 - 4(2)(3) = 1 - 24 = -23 < 0,

so there are no real roots; the parabola stays entirely above the $x$ -axis.

State the answers. (a) two irrational roots; (b) one repeated root; (c) no real roots. The discriminant settled all three without solving any equation.

Find the greatest height of a projectile in context

A ball is thrown so that its height $h$ metres after $t$ seconds is $h = -5t^2 + 20t + 25$ . Find the greatest height and when it is reached.

Recognise the model. This is a concave-down parabola ( $a = -5 < 0$ ), so the greatest height is the maximum value at the vertex.

Find the time of the maximum (the axis of symmetry).

t = -\frac{b}{2a} = -\frac{20}{2 \times (-5)} = -\frac{20}{-10} = 2.

Substitute to find the greatest height.

h = -5(2)^2 + 20(2) + 25 = -20 + 40 + 25 = 45.

State the answer. The ball reaches a greatest height of $45$ m after $2$ seconds. (The starting height, at $t = 0$ , is $h = 25$ m, the $y$ -intercept; the peak is $20$ m above that.)

Final answers: $y = x^2 - 2x - 8$ has intercepts $(-2, 0)$ , $(4, 0)$ , $(0, -8)$ and vertex $(1, -9)$ ; $y = x^2 + 6x + 1 = (x + 3)^2 - 8$ has vertex $(-3, -8)$ , range $y \ge -8$ and $x$ -intercepts $-3 \pm 2\sqrt{2}$ ; $y = 2x^2 - 12x + 7 = 2(x - 3)^2 - 11$ has vertex $(3, -11)$ ; $2x^2 + 3x - 5 = 0$ gives $x = 1$ or $x = -\dfrac{5}{2}$ ; the discriminants $13$ , $0$ , $-23$ give two, one and no real roots; and the projectile $h = -5t^2 + 20t + 25$ peaks at $45$ m after $2$ seconds.

Common traps

Sign errors when completing the square: After halving and squaring, you add and subtract the same number: $x^2 + 6x = (x + 3)^2 - 9$ . Forgetting the $-9$ changes the constant and moves the vertex. Always expand your answer mentally to check it returns the original.
Forgetting to factor the leading coefficient out first on a non-monic: For $2x^2 - 12x + 7$ you must write $2(x^2 - 6x) + 7$ before completing the square, and then multiply the $-9$ back by $2$ . Completing the square on $2x^2 - 12x$ directly does not work.
Misreading the vertex sign from $(x - h)^2 + k$: The vertex is $(h, k)$ , so $(x + 3)^2 - 8$ has vertex $(-3, -8)$ , not $(3, -8)$ : the sign flips because the form has a minus, $x - h = x - (-3)$ .

Dropping the $\pm$ in the quadratic formula or under a square root. $(x + 3)^2 = 8$ gives $x + 3 = \pm\sqrt{8}$ , so there are two solutions. A square root in this context always brings a plus-or-minus.

Confusing "no real roots" with "no solution to the problem." $\Delta < 0$ means the parabola never meets the $x$ -axis; it does not mean you made an error. State "no real roots" as the answer.

Forgetting the discriminant is fastest for counting. To answer "how many solutions", compute $\Delta = b^2 - 4ac$ and quote the case. Solving the whole equation wastes time and risks an arithmetic slip.

Keeping a physically impossible answer in a context. A negative time or a negative length is not a valid answer to a projectile or area problem. Solve the quadratic, then reject the impossible root and say why.

Exam technique

Markers reward the right tool named, the working laid out in stages, and every feature shown on a sketch. To bank the method marks:

Match the tool to the wording. Factor for intercepts, complete the square (or use $-\dfrac{b}{2a}$ ) for the vertex, and the discriminant for "how many roots". Reaching for the wrong method is where time and marks are lost.
Compute the discriminant first on any "number of roots", "equal roots" or "no real roots" question, and quote the case ( $\Delta > 0$ , $\Delta = 0$ , $\Delta < 0$ ). For a "find $k$ " version, write $\Delta$ in terms of $k$ before setting the condition.
Show all features on a sketch. Label the concavity, the $y$ -intercept, both $x$ -intercepts and the vertex with coordinates; each labelled feature is typically a mark.
State the axis formula explicitly. Write $x = -\dfrac{b}{2a}$ , substitute the numbers, then substitute that $x$ back for the vertex's $y$ -value, so the maximum or minimum value is clearly the $y$ -coordinate.
Keep exact form. Leave irrational roots as simplified surds such as $-3 \pm 2\sqrt{2}$ unless a decimal is requested; only round at the very end.
In a context, define the variable and reject impossible roots. Say what $x$ or $t$ represents, then discard a negative time or length, stating that it is not physical.
Check the vertex two ways when you have time. The completed-square vertex and the $-\dfrac{b}{2a}$ value must agree; a mismatch flags an arithmetic error.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksSketch the parabola

y = x^2 - 4

, showing the coordinates of the

x

-intercepts, the

y

-intercept and the vertex.

Show worked solution →

Read off the shape and $y$ -intercept. The coefficient of $x^2$ is $a = 1 > 0$ , so the parabola is concave up (a minimum). Putting $x = 0$ gives $y = -4$ , so the $y$ -intercept is $(0, -4)$ .

Find the $x$ -intercepts (put $y = 0$ ). Solve $x^2 - 4 = 0$ , which is a difference of two squares:

x^2 - 4 = (x - 2)(x + 2) = 0 \;\Rightarrow\; x = 2 \text{ or } x = -2.

So the $x$ -intercepts are $(-2, 0)$ and $(2, 0)$ .

Find the vertex. The axis of symmetry is halfway between the intercepts, at $x = \dfrac{-2 + 2}{2} = 0$ . Substituting $x = 0$ gives $y = -4$ , so the vertex is $(0, -4)$ (here it coincides with the $y$ -intercept).

State the sketch. A concave-up parabola with $x$ -intercepts $(-2, 0)$ and $(2, 0)$ , $y$ -intercept and vertex both at $(0, -4)$ , symmetric about the $y$ -axis. Its minimum value is $-4$ .

foundation3 marksComplete the square to write

y = x^2 - 8x + 10

in the form

y = (x - h)^2 + k

. Hence state the coordinates of the vertex, the minimum value, and the range.

Show worked solution →

Complete the square. Halve the coefficient of $x$ (which is $-8$ ) to get $-4$ , then square to get $16$ . Add and subtract $16$ :

y = (x^2 - 8x + 16) - 16 + 10 = (x - 4)^2 - 6.

Read off the vertex: In the form $y = (x - h)^2 + k$ the vertex is $(h, k)$ , so the vertex is $(4, -6)$ .
State the minimum value: Because $(x - 4)^2 \ge 0$ for all $x$ and equals $0$ only at $x = 4$ , the smallest $y$ can be is $-6$ . So the minimum value is $-6$ , reached at $x = 4$ .
State the range: Since $y$ is never below $-6$ , the range is $y \ge -6$ .

core3 marksThe equation

x^2 + mx + 9 = 0

has exactly one solution. Find the possible values of

m

Show worked solution →

Translate the condition into the discriminant. Exactly one solution (a repeated root) means the discriminant is zero, $\Delta = 0$ .

Write the discriminant. With $a = 1$ , $b = m$ and $c = 9$ ,

\Delta = b^2 - 4ac = m^2 - 4(1)(9) = m^2 - 36.

Set it to zero and solve.

m^2 - 36 = 0 \;\Rightarrow\; m^2 = 36 \;\Rightarrow\; m = 6 \text{ or } m = -6.

State the answer. The equation has exactly one solution when $m = 6$ or $m = -6$ . (Check: $m = 6$ gives $x^2 + 6x + 9 = (x + 3)^2 = 0$ , a perfect square, confirming the single repeated root.)

core4 marksSketch the parabola

y = 2x^2 + 4x - 6

, showing the

y

-intercept, the

x

-intercepts and the vertex.

Show worked solution →

Note the shape and $y$ -intercept. Here $a = 2 > 0$ , so the parabola is concave up. Putting $x = 0$ gives $y = -6$ , so the $y$ -intercept is $(0, -6)$ .

Find the $x$ -intercepts by factoring. Take out the common factor $2$ first, then factor the monic quadratic:

y = 2(x^2 + 2x - 3) = 2(x + 3)(x - 1).

Setting $y = 0$ gives $x = -3$ or $x = 1$ , so the $x$ -intercepts are $(-3, 0)$ and $(1, 0)$ .

Find the vertex. The axis of symmetry is halfway between the intercepts, at $x = \dfrac{-3 + 1}{2} = -1$ . Substituting $x = -1$ :

y = 2(-1)^2 + 4(-1) - 6 = 2 - 4 - 6 = -8,

so the vertex is $(-1, -8)$ .

State the sketch. A concave-up parabola through $(-3, 0)$ and $(1, 0)$ , $y$ -intercept $(0, -6)$ , vertex $(-1, -8)$ , symmetric about $x = -1$ . The minimum value is $-8$ .

exam5 marksA stone is thrown from the top of a cliff. Its height

h

metres above the sea

t

seconds after release is

h = -5t^2 + 10t + 15

. (a) State the height of the cliff. (b) Find the greatest height the stone reaches, and when it reaches it. (c) Find the time at which the stone hits the sea.

Show worked solution →

(a) Height of the cliff (the starting height). At $t = 0$ , $h = -5(0)^2 + 10(0) + 15 = 15$ . So the cliff is $15$ m above the sea.

(b) Greatest height (the vertex of the parabola). Here $a = -5 < 0$ , so the parabola is concave down and the vertex is a maximum. The axis of symmetry is

t = -\frac{b}{2a} = -\frac{10}{2(-5)} = 1.

Substituting $t = 1$ :

h = -5(1)^2 + 10(1) + 15 = -5 + 10 + 15 = 20.

So the greatest height is $20$ m, reached after $1$ second.

(c) When the stone hits the sea (put $h = 0$ ). Solve $-5t^2 + 10t + 15 = 0$ . Divide both sides by $-5$ :

t^2 - 2t - 3 = 0 \;\Rightarrow\; (t - 3)(t + 1) = 0 \;\Rightarrow\; t = 3 \text{ or } t = -1.

State the answer. Time cannot be negative, so reject $t = -1$ . The stone hits the sea after $3$ seconds. (Cliff height $15$ m; greatest height $20$ m at $t = 1$ s; lands at $t = 3$ s.)

exam4 marksA farmer has

60

m of fencing to enclose a rectangular paddock. (a) If one side is

x

metres, show that the area is

A = 30x - x^2

. (b) Find the dimensions that give the greatest area, and state that greatest area.

Show worked solution →

(a) Build the area function. The perimeter is $60$ m, so the two pairs of sides satisfy $2x + 2y = 60$ , giving $y = 30 - x$ for the adjacent side. The area is

A = x \times y = x(30 - x) = 30x - x^2.

(b) Find the maximum (the vertex). Write $A = -x^2 + 30x$ , so $a = -1 < 0$ and the vertex is a maximum. The axis of symmetry is

x = -\frac{b}{2a} = -\frac{30}{2(-1)} = 15.

The adjacent side is then $y = 30 - 15 = 15$ m, and the area is

A = -(15)^2 + 30(15) = -225 + 450 = 225.

State the answer. The greatest area is $225 \text{ m}^2$ , achieved when the paddock is a $15 \text{ m} \times 15 \text{ m}$ square. (A square always gives the greatest area for a fixed perimeter, which the equal sides confirm.)

What this dot point is asking

The answer

The parabola: shape, concavity and the basic features

Sketching by factoring: the x-intercepts

Completing the square: the vertex form

The axis of symmetry and the maximum or minimum value

The quadratic formula and the discriminant

Quadratics in real contexts

How exam questions ask about quadratics

Practice questions

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