Quadratic Functions Quiz — Secondary 5

Answer: −7

Since a = 2 is positive, the parabola opens upward, so its vertex is the lowest point and the minimum value is the y-coordinate of that vertex. Two steps, and you cannot skip the first. Step one, find where the vertex sits: x = −b / (2a) = −8 / (2 × 2) = −8 / 4 = −2. Step two, find how low the curve goes there, by substituting that x back into the rule: y = 2(−2)² + 8(−2) + 1 = 2(4) − 16 + 1 = 8 − 16 + 1 = −7. So the vertex is (−2, −7) and the minimum value is −7. Completing the square confirms it: 2x² + 8x + 1 = 2(x² + 4x) + 1 = 2((x + 2)² − 4) + 1 = 2(x + 2)² − 8 + 1 = 2(x + 2)² − 7. Every wrong option is a step gone astray, and none of them can be dismissed without working. The answer −2 is where the minimum happens, not the minimum itself — read the question twice, this is the most common mistake on this type. The answer −3 comes from forgetting to multiply the compensating −4 by the 2 outside the bracket, writing −4 + 1 instead of −8 + 1. And −19 comes from dropping the square when substituting, computing 2(−2) − 16 + 1 instead of 2(−2)² − 16 + 1. A useful habit: once you have a candidate, test it against a nearby point. Here f(−1) = 2 − 8 + 1 = −5, which is above −7, and f(−3) = 18 − 24 + 1 = −5 as well — symmetric around x = −2, exactly as the vertex demands.

Answer: x < −2 or x > 4

An inequality is never solved by moving terms around as if it were an equation — you have to know where the curve crosses the axis first. Start with the equation: x² − 2x − 8 = 0. Factor it, looking for two numbers with product −8 and sum −2, which are −4 and 2: (x − 4)(x + 2) = 0, so the zeros are −2 and 4. Careful with the signs here: the factors are (x − 4) and (x + 2), but the zeros are +4 and −2, the opposite signs. Getting that backwards is what produces the answers built on −4 and 2. Those two zeros cut the number line into three stretches, so test one value in each. At x = −3: 9 + 6 − 8 = 7, positive. At x = 0: −8, negative. At x = 5: 25 − 10 − 8 = 7, positive. The inequality asks for positive values, so the answer is the two outer stretches: x < −2 or x > 4. The shape explains it without any testing: a = 1 is positive, the parabola opens upward, so it is below the axis between its zeros and above the axis outside them. Remember that pattern — greater than zero means outside for an upward parabola, less than zero means between — and note that the inequality is strict, so the zeros themselves are excluded.

Answer: ]−∞, 5]

The range is the set of all the heights the function actually reaches. Two things decide it, and you need both. First, which way does the curve open? The coefficient in front of the square is a = −2, which is negative, so the parabola opens downward: it has a highest point, nothing above it, and it falls away forever below. That already tells you the range is bounded above and unbounded below, which is why it is written with −∞ on the left. Second, how high is that highest point? The function is already in vertex form a(x − h)² + k, so the vertex is (1, 5) and the maximum height is k = 5. Put the two together: the function reaches 5 and everything below it, so the range is ]−∞, 5]. The bracket closes on 5 because that value is actually attained, at x = 1. Check with a value: at x = 3, f(3) = −2(4) + 5 = −3, below 5; and no x can push the output above 5, since −2(x − 1)² is never positive. Each wrong option mistakes one ingredient for another. [5, +∞[ has the right vertex but the wrong direction, which is what happens if you overlook the minus sign in front of the 2. ]−∞, 1] uses the x-coordinate of the vertex instead of the y-coordinate — the most common slip of all, since the 1 is right there in the bracket. And ]−∞, −2] uses the coefficient a as if it were a height. Remember the shape of the answer as well: for a parabola the range is always a half-line bounded at k, never a bounded interval and never all of the real numbers.

Answer: y = x² − 2x − 3

Read the graph in three steps, and this time the third one is what decides. First, the zeros: the curve crosses the x-axis at x = −1 and x = 3, so the rule has the form y = a(x + 1)(x − 3). Second, the y-intercept: the curve crosses the vertical axis at −3. Substitute x = 0 into the factored form: a(1)(−3) = −3, which gives a = 1. So y = (x + 1)(x − 3) = x² − 3x + x − 3 = x² − 2x − 3, the first option. Third, confirm on your own: the axis of symmetry runs halfway between the zeros, at x = 1, and the rule puts the vertex there at 1 − 2 − 3 = −4, which is where the drawing bottoms out. It fits. Now why the other three fail, because none of them can be dismissed at a glance — all four open upward and all four cross the x-axis. The rule y = x² + 2x − 3 factors as (x + 3)(x − 1), so its zeros are −3 and 1: it is the mirror image of the drawing, and it even shares the same y-intercept of −3, which is exactly why reading the intercept alone is not enough. The rule y = x² − 2x − 8 factors as (x − 4)(x + 2), zeros −2 and 4, and its y-intercept is −8, far too low. And y = 2x² − 4x − 6 factors as 2(x + 1)(x − 3) — it has the very same zeros, −1 and 3, as the correct answer. The only thing that separates them is the stretch: its y-intercept is −6 instead of −3 and its vertex is (1, −8) instead of (1, −4), so the curve would be twice as deep. The lesson is that two zeros never determine a parabola on their own; you always need a third point to pin down a.

Answer: It has no real zeros.

All four statements are claims about the zeros, and not one of them can be settled by looking at the shape of the curve — you have to compute the discriminant. Δ = b² − 4ac, and with a = 2, b = 3 and c = 5 that gives Δ = 9 − 4(2)(5) = 9 − 40 = −31. A negative discriminant means the quadratic formula would ask for the square root of a negative number, which has no real value, so the function has no real zeros: the third statement is the true one. The three cases are worth memorising: Δ > 0 gives two zeros, Δ = 0 gives exactly one and the vertex sits on the axis, and Δ < 0 gives none. That settles the first two statements at once. The fourth is the interesting one, because it is a finer claim: zeros are irrational when Δ is positive but not a perfect square — for instance x² − 2 = 0 has Δ = 8 and zeros ±√2. It is a statement you can only judge after computing Δ, and here Δ is negative, so there are no zeros at all and the claim collapses. Geometrically the whole picture is simple: a = 2 is positive so the parabola opens upward, and its vertex sits at x = −3/4 with y = 2(9/16) − 9/4 + 5 = 9/8 − 18/8 + 40/8 = 31/8, comfortably above the x-axis, which the curve therefore never touches.

Answer: 4

The formula is x = (−b ± √(b² − 4ac)) / (2a). Identify the coefficients with their signs: a = 2, b = −5, c = −12. Discriminant: Δ = (−5)² − 4(2)(−12) = 25 + 96 = 121. Watch that step — minus four times two times negative twelve is plus ninety-six, because two negatives make a positive. The square root of 121 is 11. Now the denominator is 2a = 4, not 2: with a leading coefficient other than 1, forgetting to double a is the single most common wreck. So x = (5 ± 11) / 4. With the plus, x = 16 / 4 = 4. With the minus, x = −6 / 4 = −1.5. The larger solution is 4. Check it: 2(16) − 5(4) − 12 = 32 − 20 − 12 = 0. Correct. And check the other one too: 2(2.25) + 7.5 − 12 = 4.5 + 7.5 − 12 = 0. Both roots hold.

Answer: 21 m

Because a = −5 is negative, the parabola opens downward and its vertex is a maximum — which is exactly the highest point of the throw. Find the time first: t = −b / (2a) = −20 / (2 × −5) = −20 / −10 = 2 seconds. Then find the height at that moment by substituting: h = −5(2)² + 20(2) + 1 = −5(4) + 40 + 1 = −20 + 41 = 21 metres. So the ball peaks at 21 m, two seconds after being thrown. Note what each number means physically: the constant 1 is the height at t = 0, so the ball left a hand one metre off the ground, and the −5 comes from gravity. And be careful what the question asks — the answer is the height 21, not the time 2. Mixing up the two coordinates of the vertex is the most common mistake in optimisation problems.

Answer: 12

This is the discriminant read backwards, with a leading coefficient that is not 1. An equation ax² + bx + c = 0 has exactly one solution — a double root — when Δ = b² − 4ac equals zero. Here a = 4, b = k and c = 9, so Δ = k² − 4(4)(9) = k² − 144. Set it to zero: k² = 144, so k = 12 or k = −12. Both values genuinely give a double root, which is why the question asks for the positive one: k = 12. Check it: 4x² + 12x + 9 is the perfect square (2x + 3)², whose only solution is x = −3/2. Two things are worth keeping. First, squaring removes the sign, so an equation like k² = 144 always has two answers and you must go back to the wording to choose — here the word positive does that. Second, notice how the double root shows up geometrically: the parabola touches the x-axis at a single point instead of crossing it. If k were bigger than 12 the discriminant would turn positive and the curve would cut the axis twice; if k were between −12 and 12 the discriminant would be negative and the curve would float clear of the axis.

Answer: y = 3(x − 2)² − 5

When the coefficient of x² is not 1, factor it out of the x terms before completing the square — this is the step that trips most people. Start with y = 3x² − 12x + 7 and pull the 3 out of the first two terms only: y = 3(x² − 4x) + 7. Now complete the square inside the bracket. Take the coefficient of x, which is −4, halve it to get −2, and square that to get 4. Add and subtract 4 inside the bracket: y = 3(x² − 4x + 4 − 4) + 7 = 3((x − 2)² − 4) + 7. Distribute the 3 over both pieces — the second one too, and this is the other classic slip: y = 3(x − 2)² − 12 + 7 = 3(x − 2)² − 5. So the vertex is (2, −5). Expand backwards to verify: 3(x − 2)² − 5 = 3(x² − 4x + 4) − 5 = 3x² − 12x + 12 − 5 = 3x² − 12x + 7. Correct. The wrong options map the two standard mistakes: y = 3(x − 2)² + 7 keeps the original constant without paying for the 4 that was added inside, and y = (3x − 6)² − 5 squares the whole 3x, which expands to 9x² − 36x + 31 — a different function entirely. And y = 3(x − 6)² − 101 halves nothing, using −12 where −4 belongs.

Answer: 3

You are given the vertex form with one unknown left, and one point the curve has to pass through. Saying that the point (0, 7) is on the parabola means its coordinates make the rule true, so substitute x = 0 and y = 7: 7 = a(0 − 2)² − 5. Work out the bracket first, before anything else: (0 − 2)² = (−2)² = 4, and the square makes it positive. So 7 = 4a − 5. Add 5 to both sides: 12 = 4a. Divide by 4: a = 3. The full rule is y = 3(x − 2)² − 5, with vertex (2, −5). Check by substituting the point back in: 3(0 − 2)² − 5 = 3(4) − 5 = 12 − 5 = 7. Correct. The usual slip is writing (0 − 2)² as −4, forgetting that squaring a negative gives a positive; that would hand you a = 1/2 instead. And notice what a tells you: since a = 3 is positive and greater than 1, the parabola opens upward and is three times narrower than the basic y = x².

Answer: 2

At a point of intersection the two curves share the same x and the same y, so the two expressions for y must be equal: x² + 1 = 3x − 1. Bring everything to one side to get a quadratic equal to zero, because that is the only form the zero product rule accepts. Subtract 3x and add 1: x² − 3x + 2 = 0. Factor it — two numbers with product 2 and sum −3, which are −1 and −2: (x − 1)(x − 2) = 0, so x = 1 or x = 2. The larger x-coordinate is 2. Find the full points if you want them: at x = 1, y = 1² + 1 = 2 and also 3(1) − 1 = 2, so (1, 2); at x = 2, y = 5 from both rules, so (2, 5). Both checks agree, which is the point of doing them in the second equation rather than the first. One thing worth understanding: the number of intersections is decided by the discriminant of the combined equation, here (−3)² − 4(1)(2) = 1, positive, hence two points. Had it been zero the line would be tangent to the parabola, and negative would mean they never meet.

Answer: 100 m²

Turn the situation into a quadratic function of one variable. Let x be the width. The perimeter is 40, so the two widths and two lengths together use 40 m: 2x + 2L = 40, which gives L = 20 − x. The area is then A = x(20 − x) = 20x − x². This is a quadratic with a = −1 and b = 20, and since a is negative it has a maximum at its vertex. The vertex is at x = −b / (2a) = −20 / (2 × −1) = 10. So the width is 10 m and the length is 20 − 10 = 10 m — a square. The area is A = 10 × 10 = 100 m². Check against another shape: 8 by 12 also has perimeter 40 but an area of only 96 m², which is less, as predicted. The general result is worth remembering: for a fixed perimeter, the rectangle of greatest area is always the square.