Secondary 4 Quadratic Functions Worksheet

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Finding the Rule of a Quadratic Function

   A quadratic function has its vertex at $(3,-8)$ and passes through the point $(5,0)$. Write its rule in standard (vertex) form, then expand it into general form.

2. Q2Forms of the Equation of a Quadratic Function

   The function $f(x)=2x^2-12x+16$ is given in general form.

   1. Rewrite it in standard (vertex) form and give the coordinates of the vertex.
   2. Rewrite it in factored form and give the zeroes.
3. Q3Graphing a Quadratic Function

   Sketch $f(x)=-(x+1{)}^2+4$ on the grid. State the vertex, the direction of opening, the zeroes and the $y$-intercept.

   A blank Cartesian grid for this question is on the printable PDF.

4. Q4Solving Problems Involving Quadratic Functions

   A diver leaves a platform and her height above the water, in metres, $t$ seconds later is $h(t)=-5t^2+6t+8$.

   1. What is the greatest height she reaches, and when?
   2. After how many seconds does she enter the water?
5. Q5The Inverse of Quadratic Functions

   Let $f(x)=(x-3{)}^2-5$, with domain $ℝ$.

   1. Find the rule of the inverse relation $f^{-1}$.
   2. Explain why this relation is not a function.
   3. Restrict the domain of $f$ so that the inverse is a function, and give the rule and the domain of that inverse.
6. Q6The Properties of Quadratic Functions

   Consider $f(x)=-2(x-3{)}^2+8$.

   1. Give the coordinates of the vertex and state whether $f$ has a maximum or a minimum.
   2. State the domain and the range of $f$.
   3. Over which interval is $f$ increasing?
   4. Find the zeroes and the $y$-intercept of $f$.
7. Q7The Quadratic Formula

   Solve $3x^2-7x-6=0$ using the quadratic formula. Show the value of the discriminant.

8. Q8The Quadratic Function

   Only some of the rules below define quadratic (second-degree polynomial) functions. Identify them and justify each decision by simplifying the rule.

   1. $f(x)=3(x-2{)}^2-(3x^2+1)$
   2. $g(x)=x(x-5)+7$
   3. $h(x)={\displaystyle \frac{2}{x^2}}+x$
   4. $k(x)=(x+1{)}^2-(x-1{)}^2$
9. Q9The Role of Parameters in a Quadratic Function

   For a quadratic function written $f(x)=a(x-h{)}^2+k$, state a rule that gives the number of zeroes using only the signs of $a$ and $k$, and justify it algebraically. Then apply your rule to $f(x)=-3(x+2{)}^2-5$ without computing any zero.

10. Q10The Zeroes of a Quadratic Function

    Find the zeroes of $f(x)=2x^2-11x+12$ by factoring.

11. Q11Synthesis — drawing on several sheets in this topic

    A cycling club rents a chalet and sells weekend passes. Market research shows that the number of passes sold is a first-degree function of the price: at $60 a pass, $240$ passes sell; at $80 a pass, $200$ passes sell. The revenue is the price multiplied by the number of passes sold.

    1. Find the rule $n(p)$ giving the number of passes sold at a price of $p$ dollars.
    2. Express the revenue $R(p)$ and write it in standard form.
    3. Which price maximizes the revenue, and what is that maximum revenue?
    4. The club must raise at least $14{,}400. For which prices is this achieved?

The 9 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Three forms of the same parabola, and what each one hands you

Almost everything on this sheet is a decision about which form to write the rule in. The three forms describe the same curve; they differ only in what you can read off them without doing any work. Choosing badly turns a two-line question into a page.

One thing the tree does not say: not every quadratic has a factored form over the reals. If the parabola never meets the x-axis there are no zeroes to put in brackets. The general and standard forms always exist; the factored form is the conditional one.

Completing the square when a ≠ 1

With a = 1 the move is mechanical: halve the coefficient of x, square it, add and subtract it. Q2(a) has a leading coefficient in front, and that changes the order of operations completely.

Reading a parabola: vertex, opening, range, variation

Q3 and Q6 are the same reading, one with a sketch and one without. From f(x) = a(x − h)² + k:

• Vertex (h, k) — and watch the sign, because (x + 1)² means h = −1. The bracket must read (x − h) before you read h out of it.
• Opening from the sign of a: upward when a > 0, downward when a < 0. That single sign then decides the next two lines.
• Range: [k, +∞[ if it opens upward, ]−∞, k] if it opens downward. The domain of a quadratic function is always ℝ unless a context restricts it.
• Variation: the split is always at x = h. A downward parabola climbs to its vertex and then falls, so it is increasing on ]−∞, h] — which feels backwards to students who memorised the upward case.
• Zeroes: from standard form, isolate the square and take the square root of both sides, remembering both signs. (x − h)² = m gives x = h ± √m. The quadratic formula is unnecessary here.
• y-intercept: substitute x = 0. One arithmetic line, and it is the point students forget to plot.

Building the rule from what you are told

Q1 gives you a vertex and one other point. That combination is a signal: write a(x − h)² + k straight away with h and k filled in, substitute the other point, and solve for the one unknown left. Two points and no vertex would not be enough — a parabola needs three pieces of information, and a vertex is worth two of them.

The same logic runs the other way. If you are given two zeroes and one more point, start from a(x − x₁)(x − x₂). If you are given three scattered points, you are back to general form and a system of three equations. Pick the form that makes the information you have free.

Zeroes: how many, before you find them

Q7 asks for the discriminant explicitly, and Q9 asks you to count zeroes without computing any. They are two views of the same fact.

From standard form you do not need Δ at all. Solving a(x − h)² + k = 0 gives (x − h)² = −k⁄a, and a square is never negative. So: a and k of opposite signs gives two zeroes, k = 0 gives exactly one (the vertex sits on the axis), and a and k of the same sign gives none — the whole parabola is stranded on one side of the axis. Q9 asks you to state and justify that rule, so the argument above, not the conclusion, is what earns the marks.

The inverse of a parabola, and why it needs a restriction

Q5 is the conceptual question on this sheet. Swap x and y in y = a(x − h)² + k and isolate:

That ± is the entire answer to part (b). It says one value of x is paired with two values of y, which a function is not allowed to do. The reason sits in the original parabola: it is symmetric about its axis, so two different inputs share every output except the vertex — it fails the horizontal line test.

Part (c) fixes it. Cut the parabola at its vertex and keep one branch: ]−∞, h] or [h, +∞[. On one branch the function is strictly monotonic, so it is injective, so the inverse is a function — and it keeps only one sign of the radical: the + branch for the increasing side, the − branch for the decreasing side. Then state the domain of the inverse, which is the range of the restricted f, and is where the marks usually go missing.

Is it even quadratic? (Q8)

Four rules, and only some of them are second-degree. The rule of the game is that degree can only be read from a simplified expression. Expand and collect first; the x² terms may cancel completely, and a rule that looks quadratic then turns out to be first-degree. A rule with the variable in a denominator or under a radical is not a polynomial at all, and the justification the question wants is the simplification, written out — not the verdict.

Contexts: Q4 and the synthesis question

Q4 gives a height in general form. The vertex is at t = −b⁄(2a), and substituting that back gives the maximum height; "when does she enter the water" is a zero. Two habits earn the marks here: answer with units, and say why you rejected a root. A negative time is not merely wrong, it is outside the domain of the model, and one clause saying so is what the marker is looking for.

The last question builds a quadratic from scratch. The number of items sold is a first-degree function of the price, revenue is price × quantity, and multiplying those two first-degree expressions is what produces the parabola. Once it is in standard form the maximising price is the h and the maximum revenue is the k. The final part is a quadratic inequality: because the parabola opens downward, "revenue at least this much" is satisfied between the two solutions, so the answer is a closed interval, not a single price. Sketching the parabola and marking the horizontal line at that revenue makes the direction obvious and takes ten seconds.

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Getting the most out of it

Rewrite every rule in all three forms

Take the rules in Q2, Q3, Q6 and Q10 and, whatever the question asked, write each one in general, standard and factored form. Four questions become twelve conversions, and the conversions are what the exam actually tests.

Sketch first, compute second

A rough parabola with the vertex marked and the opening right tells you the number of zeroes, the range and the sign before you touch the algebra. When the algebra disagrees with the sketch, one of them is wrong and you have caught it early instead of at the bottom of the page.

Justify the root you throw away

On the context questions, get into the habit of writing the sentence: "rejected because time cannot be negative", "rejected because a width must be positive". It is a mark on almost every modelling question and it is the one students leave on the table.

Common questions