Answer key · Unit 6

The quadratic function.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 15 minutes

The shape and sign of a

When the parabola opens upward and when it opens downward, what makes it narrow or wide, and what happens when the coefficient a passes through zero.

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The 4 steps and their answers

Step 1 of 4
Open this step
The basic parabola
No practice question in this step.
The Scene at this step
a
1
y = x² / y = ax²
y = ax²default
Step 2 of 4
Open this step
Through zero
Practice · Predict, then look
Question
Right now $a = 1$ ⁠. You are going to lower a step by step to $a = -1$ ⁠, and on the way it will pass through zero. What happens to the parabola along the way?

Correct answer
- Correct It gets wider and wider, becomes a straight line at $a = 0$ ⁠, and then opens downward
Setup, choices and feedback
The Scene at this step
a
1fixed
y = x² / y = ax²
y = ax²default
Choices
- Correct It gets wider and wider, becomes a straight line at $a = 0$ ⁠, and then opens downward
- Wrong It flips over all at once and keeps the same width
  Misconception: Thinks the parabola flips from opening up to opening down in one jump as a passes zero, keeping its width.
  Feedback from the catalogue
  The coefficient $a$ ⁠ changes smoothly, and so does the parabola: as $a$ ⁠ gets closer to zero the parabola grows wider and wider, at $a = 0$ ⁠ only a straight line is left, and only after that does it open downward.
- Wrong It gets narrower and narrower, then flips over
  Misconception: Thinks a larger |a| makes the parabola wider.
  Feedback from the catalogue
  When $|a|$ ⁠ is large, $y$ ⁠ grows faster with every step in $x$ ⁠, so the arms climb quickly and the parabola is narrower. For example, at $x = 1$ ⁠: $y = 3$ ⁠ for $3x^2$ ⁠, but only $y = 0.5$ ⁠ for $0.5x^2$ ⁠.
- Wrong It keeps opening upward, only wider
  Feedback
  Try one value: at $a = -1$ ⁠ and $x = 1$ ⁠ you get $y = -1$ ⁠, below the axis. The arms are already heading down.
Once the prediction is locked in, the Scene changes
a
-1fixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
When $a$ ⁠ is small, $ax^2$ ⁠ grows slowly and the arms are flat. At $a = 0$ ⁠ the $x^2$ ⁠ term disappears and you are left with $y = 0$ ⁠, a straight line. Past zero every $y$ ⁠ changes sign, and the parabola opens downward. Move a slowly and watch the change happen.

Worked solution
Follow one point, $x = 2$ ⁠, where $y = a \cdot 2^2 = 4a$ ⁠:
- At $a = 1$ ⁠: $y = 4$ ⁠.
- At $a = 0.5$ ⁠: $y = 2$ ⁠. The arm is lower, so the parabola is wider.
- At $a = 0$ ⁠: $y = 0$ ⁠, and the same for every $x$ ⁠: the graph is the straight line $y = 0$ ⁠.
- At $a = -0.5$ ⁠: $y = -2$ ⁠, and at $a = -1$ ⁠: $y = -4$ ⁠. The arms go down, and the parabola opens downwards.
Every value of $y$ ⁠ changes smoothly with $a$ ⁠, so there is no sudden flip: the parabola flattens, becomes a straight line and opens the other way.
Step 3 of 4
Open this step
Up or down
Practice · Choose
Question
Which way does the parabola $y = -2x^2 + 5$ ⁠ open?

Correct answer
- Correct Downward
Setup, choices and feedback
The Scene at this step
a
1
y = x² / y = ax²
y = ax²default
Choices
Shown in a random order.
- Correct Downward
- Wrong Upward, because $5$ ⁠ is positive
  Misconception: Thinks the sign of the constant term decides whether the parabola opens up or down.
  Feedback from the catalogue
  The constant term only lifts or lowers the graph. The direction is set by the sign of $a$ ⁠, the coefficient of $x^2$ ⁠: when $x$ ⁠ is large, the term $ax^2$ ⁠ outweighs everything else.
- Wrong You cannot tell without drawing it
  Feedback
  The coefficient of $x^2$ ⁠ is enough on its own: its sign wins out once $x$ ⁠ is large.
Feedback for a right answer
The coefficient of $x^2$ ⁠ is $a = -2$ ⁠, which is negative, so the parabola opens downward and its vertex is a maximum point. The $5$ ⁠ only lifts the vertex to $y = 5$ ⁠.

Worked solution
Look at the coefficient of $x^2$ ⁠: here $a = -2$ ⁠, negative. As $x$ ⁠ moves away from zero, the term $-2x^2$ ⁠ grows in size much faster than any constant, so it decides the direction. For example, at $x = 10$ ⁠: $y = -2 \cdot 100 + 5 = -195$ ⁠.

The arms go down on both sides, so the parabola opens downwards. The $5$ ⁠ sets only the height of the vertex $(0, 5)$ ⁠, which is the maximum point.
Step 4 of 4
Open this step
Narrow or wide
Practice · Choose
Question
Which parabola is narrower: $y = 3x^2$ ⁠ or $y = 0.5x^2$ ⁠?

Correct answer
- Correct $y = 3x^2$ ⁠
Setup, choices and feedback
The Scene at this step
a
0.5
y = x² / y = ax²
y = ax²default
Choices
Shown in a random order.
- Correct $y = 3x^2$ ⁠
- Wrong $y = 0.5x^2$ ⁠
  Misconception: Thinks a larger |a| makes the parabola wider.
  Feedback from the catalogue
  When $|a|$ ⁠ is large, $y$ ⁠ grows faster with every step in $x$ ⁠, so the arms climb quickly and the parabola is narrower. For example, at $x = 1$ ⁠: $y = 3$ ⁠ for $3x^2$ ⁠, but only $y = 0.5$ ⁠ for $0.5x^2$ ⁠.
- Wrong They are the same width; only the height is different
  Feedback
  Both pass through $(0, 0)$ ⁠, but at $x = 1$ ⁠ one is already at $y = 3$ ⁠ and the other only at $y = 0.5$ ⁠. The arms do not rise at the same angle.
Feedback for a right answer
For $y = 3x^2$ ⁠ every $y$ ⁠ is $6$ ⁠ times what it is for $y = 0.5x^2$ ⁠, so the arms climb faster and the parabola is narrower. The larger $|a|$ ⁠ is, the narrower the parabola; the closer it is to zero, the wider.

Worked solution
Compare the two parabolas at the same $x$ ⁠:
- At $x = 1$ ⁠: $3x^2 = 3$ ⁠, but $0.5x^2 = 0.5$ ⁠.
- At $x = 2$ ⁠: $3x^2 = 12$ ⁠, but $0.5x^2 = 2$ ⁠.
$y = 3x^2$ ⁠ climbs faster. Another way to see it: both reach the height $y = 3$ ⁠, but $3x^2 = 3$ ⁠ already at $x = \pm 1$ ⁠, while $0.5x^2 = 3$ ⁠ only at $x = \pm\sqrt{6} \approx \pm 2.45$ ⁠. At the same height the first is $2$ ⁠ wide and the second almost $5$ ⁠, so $y = 3x^2$ ⁠ is narrower.

Lesson 2 of 4 · About 20 minutes

Shifting a parabola

How to move y = x² right, left, up and down, and how to read the vertex straight off vertex form.

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The 4 steps and their answers

Step 1 of 4
Open this step
Three steps from the basic parabola
No practice question in this step.
The Scene at this step
a
2
Horizontal shift p
3
Vertical shift k
-2
y = x² / The whole curve
y = x²
Step 2 of 4
Open this step
Which way?
Practice · Choose
Question
The graph of $y = (x-3)^2$ ⁠ is the graph of $y = x^2$ ⁠ after a shift. Which shift?

Correct answer
- Correct $3$ ⁠ units to the right
Setup, choices and feedback
The Scene at this step
a
1
Horizontal shift p
0
Vertical shift k
0
y = x² / The whole curve
Up or down
Choices
Shown in a random order.
- Correct $3$ ⁠ units to the right
- Wrong $3$ ⁠ units to the left
  Misconception: Reads the horizontal shift with the wrong sign: (x − 3)² moved left, or the vertex of (x + 1)² at x = 1.
  Feedback from the catalogue
  Inside the brackets the sign works the opposite way to how it looks: $(x-3)^2$ ⁠ is zero when $x = 3$ ⁠, so the vertex moves right, to $x = 3$ ⁠. To check, substitute the vertex’s $x$ ⁠ and make sure the brackets come to zero.
- Wrong $3$ ⁠ units down
  Feedback
  The change is inside the brackets, together with $x$ ⁠, so it moves the graph sideways. A shift down is a number subtracted outside the brackets, as in $x^2 - 3$ ⁠.
- Wrong $3$ ⁠ units up
  Feedback
  The change is inside the brackets, together with $x$ ⁠, so it moves the graph sideways, not up or down.
Feedback for a right answer
$(x-3)^2$ ⁠ is zero when $x = 3$ ⁠, so the vertex moves from $x = 0$ ⁠ to $x = 3$ ⁠: a shift to the right. Every point of $y = x^2$ ⁠ now sits $3$ ⁠ units to the right of where it was.

Worked solution
Find the vertex: $(x-3)^2$ ⁠ is a square, so it is smallest, zero, when $x - 3 = 0$ ⁠, that is at $x = 3$ ⁠. The vertex has moved from $(0, 0)$ ⁠ to $(3, 0)$ ⁠.

The same happens to every point: the value $y = x^2$ ⁠ takes at $x = 1$ ⁠ now comes at $x = 4$ ⁠, because $(4-3)^2 = 1^2$ ⁠. The whole graph has moved $3$ ⁠ units to the right. A minus inside the brackets moves it right, because you need a bigger $x$ ⁠ to reach the same value.
Step 3 of 4
Open this step
Hit the ring
Practice · Reach the target
Question
Move the parabola so that its vertex lands on the ring at $(2, -3)$ ⁠, then press Check.

Goal
- 2 ± 0.01
  At the start of the step: 0
- -3 ± 0.01
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
a
1fixed
Horizontal shift p
0
Vertical shift k
0
y = x² / The whole curve
Up or down
Other settings
Target/nodes/graph/props/target = [2,-3]
Read from the Scene at
X of the vertex; NaN when a = 0/nodes/graph/outputs/vertexX
Read from the Scene at
Y of the vertex; NaN when a = 0/nodes/graph/outputs/vertexY
One setting that reaches it
Horizontal shift p
2
Vertical shift k
-3
Feedback for a right answer
The vertex is at $(2, -3)$ ⁠, and the function is $y = (x-2)^2 - 3$ ⁠. Watch the signs: $2$ ⁠ to the right is written $x - 2$ ⁠ inside the brackets, and $3$ ⁠ down is written $-3$ ⁠ outside them.

Feedback for a wrong answer
Not there yet. Look at the pink vertex: it has to move $2$ ⁠ to the right and $3$ ⁠ down from the origin.

Hint
Horizontal shift p moves the vertex sideways, and Vertical shift k moves it up and down. The vertex of $y = (x-p)^2 + k$ ⁠ is $(p, k)$ ⁠.

Worked solution
In vertex form $y = (x-p)^2 + k$ ⁠ the vertex is $(p, k)$ ⁠. You need the vertex at $(2, -3)$ ⁠, so:
1. Set Horizontal shift p to $2$ ⁠: the vertex moves $2$ ⁠ units to the right.
2. Set Vertical shift k to $-3$ ⁠: the vertex moves $3$ ⁠ units down.
The function is $y = (x-2)^2 - 3$ ⁠. Check: at $x = 2$ ⁠ the brackets are zero and $y = -3$ ⁠.
Step 4 of 4
Open this step
Read the vertex
Practice · Choose
Question
What is the vertex of the parabola $y = (x+1)^2 - 4$ ⁠?

Correct answer
- Correct $(-1, -4)$ ⁠
Setup, choices and feedback
The Scene at this step
a
1
Horizontal shift p
2
Vertical shift k
1
y = x² / The whole curve
Up or down
Choices
Shown in a random order.
- Correct $(-1, -4)$ ⁠
- Wrong $(1, -4)$ ⁠
  Misconception: Reads the horizontal shift with the wrong sign: (x − 3)² moved left, or the vertex of (x + 1)² at x = 1.
  Feedback from the catalogue
  Inside the brackets the sign works the opposite way to how it looks: $(x-3)^2$ ⁠ is zero when $x = 3$ ⁠, so the vertex moves right, to $x = 3$ ⁠. To check, substitute the vertex’s $x$ ⁠ and make sure the brackets come to zero.
- Wrong $(-4, -1)$ ⁠
  Feedback
  The pair is written $(x, y)$ ⁠: first the sideways shift, from inside the brackets, then the height, from outside them.
- Wrong $(-1, 4)$ ⁠
  Feedback
  The number outside the brackets is read just as it is written: $-4$ ⁠ lowers the vertex by $4$ ⁠ units.
Feedback for a right answer
$(x+1)^2$ ⁠ is zero when $x = -1$ ⁠, and there $y = -4$ ⁠. Check it in the Scene: set Horizontal shift p to $-1$ ⁠ and Vertical shift k to $-4$ ⁠.

Worked solution
Write the function as $y = (x-p)^2 + k$ ⁠: $(x+1)^2 = (x-(-1))^2$ ⁠, so $p = -1$ ⁠ and $k = -4$ ⁠. The vertex is $(p, k) = (-1, -4)$ ⁠.

Check: at $x = -1$ ⁠ the brackets are zero and $y = 0 - 4 = -4$ ⁠. A square cannot be negative, so this is the smallest value $y$ ⁠ takes.

Lesson 3 of 4 · About 20 minutes

Roots and the discriminant

Where the parabola crosses the x-axis, and how a single number, the discriminant, tells you in advance whether there are two roots, one or none.

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The 4 steps and their answers

Step 1 of 4
Open this step
Where the graph meets the axis
No practice question in this step.
The Scene at this step
a
1
b
-2
c
-3
The curve / Roots and Δ
Roots and Δ
Step 2 of 4
Open this step
Predict before you look
Practice · Predict, then look
Question
The parabola $y = x^2 - 2x - 3$ ⁠ has two roots. Now raise c from $-3$ ⁠ to $1$ ⁠, which lifts the whole graph $4$ ⁠ units. How many roots will $y = x^2 - 2x + 1$ ⁠ have?

Correct answer
- Correct One
Setup, choices and feedback
The Scene at this step
a
1fixed
b
-2fixed
c
-3fixed
The curve / Roots and Δ
Roots and Δ
Choices
- Wrong Two
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Correct One
- Wrong None
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
c
1fixed
The Scene then shows: 1 · Real roots: 0, 1 or 2 (a line: 1, or 0 when flat)/nodes/graph/outputs/rootCount
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$\Delta = (-2)^2 - 4 \cdot 1 \cdot 1 = 0$ ⁠. The vertex was at $(1, -4)$ ⁠, and lifting it by $4$ ⁠ puts it exactly on the $x$ ⁠-axis: the parabola touches the axis at a single point, $x = 1$ ⁠. That is a double root: $y = (x-1)^2$ ⁠.

Hint
Work out $\Delta = b^2 - 4ac$ ⁠ with $c = 1$ ⁠, or find the vertex: how many units below the axis is it right now?

Worked solution
After the lift the function is $y = x^2 - 2x + 1$ ⁠, with $a = 1$ ⁠, $b = -2$ ⁠ and $c = 1$ ⁠.
1. The discriminant: $\Delta = b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 1 = 4 - 4 = 0$ ⁠.
2. $\Delta = 0$ ⁠ means one (double) root: $x = -\dfrac{b}{2a} = 1$ ⁠.
Another way: the vertex of $y = x^2 - 2x - 3$ ⁠ is $(1, -4)$ ⁠, four units below the axis. Lifting by $4$ ⁠ brings it exactly onto the axis, and the parabola touches it at one point. Indeed $x^2 - 2x + 1 = (x-1)^2$ ⁠.
Step 3 of 4
Open this step
Work out the discriminant
Practice · Calculate
Question
Work out the discriminant $\Delta$ ⁠ of $y = 2x^2 + 3x - 5$ ⁠.

Correct answer
49 ± 0.01
Setup, choices and feedback
The Scene at this step
a
2fixed
b
3fixed
c
-5fixed
The curve / Roots and Δ
The curvehidden
Accepted within
± 0.01
Computed from the Scene at this step’s setting
B² − 4ac/nodes/graph/outputs/discriminant
Wrong answers it recognises
- -31 ± 0.01
  Misconception: Computes the discriminant as b² + 4ac.
  Feedback from the catalogue
  The discriminant is $\Delta = b^2 - 4ac$ ⁠, with a minus. It is what sits under the square root in the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ⁠.
Feedback for a right answer
$\Delta = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$ ⁠. It is positive, so there are two roots: $x = \dfrac{-3 \pm 7}{4}$ ⁠, that is, $x = 1$ ⁠ and $x = -2.5$ ⁠. Choose Roots and Δ under Look inside to check.

Hint
$a = 2$ ⁠, $b = 3$ ⁠, $c = -5$ ⁠. Substitute them into $\Delta = b^2 - 4ac$ ⁠, and mind the sign of $c$ ⁠.

Worked solution
1. Identify the coefficients: $a = 2$ ⁠, $b = 3$ ⁠ and $c = -5$ ⁠.
2. Substitute: $\Delta = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5)$ ⁠.
3. Mind the sign: $-4 \cdot 2 \cdot (-5) = +40$ ⁠, so $\Delta = 9 + 40 = 49$ ⁠.
If you got $-31$ ⁠, you wrote $b^2 + 4ac$ ⁠ or lost the minus sign of $c$ ⁠. $\Delta = 49$ ⁠ is positive, so the parabola has two roots: $x = \dfrac{-3 \pm \sqrt{49}}{4}$ ⁠, that is $x = 1$ ⁠ and $x = -2.5$ ⁠.
Step 4 of 4
Open this step
When the discriminant is negative
Practice · Choose
Question
The parabola $y = x^2 + 4x + 5$ ⁠ has $\Delta = -4$ ⁠. What does that tell you about its roots?

Correct answer
- Correct It has no roots: it never crosses the $x$ ⁠-axis
Setup, choices and feedback
The Scene at this step
a
1
b
4
c
5
The curve / Roots and Δ
Roots and Δ
Choices
Shown in a random order.
- Correct It has no roots: it never crosses the $x$ ⁠-axis
- Wrong Both of its roots are negative
  Misconception: Thinks a negative discriminant means the roots are negative numbers.
  Feedback from the catalogue
  The sign of $\Delta$ ⁠ does not tell you whether the roots are positive or negative, but how many there are: a negative number has no real square root, so when $\Delta < 0$ ⁠ the parabola has no real roots. It never crosses the $x$ ⁠-axis.
- Wrong It has one root
  Feedback
  There is one (double) root only when $\Delta = 0$ ⁠ exactly, which is when the vertex sits on the axis.
Feedback for a right answer
The quadratic formula contains $\sqrt{\Delta}$ ⁠, and a negative number has no real square root. On the graph, the vertex $(-2, 1)$ ⁠ is above the $x$ ⁠-axis and the parabola opens upward, so it never comes down to the axis.

Worked solution
$\Delta = 4^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4$ ⁠. The quadratic formula $x = \dfrac{-b \pm \sqrt{\Delta}}{2a}$ ⁠ needs $\sqrt{-4}$ ⁠, and a negative number has no real square root, so there are no real roots.

Completing the square shows it directly: $x^2 + 4x + 5 = (x+2)^2 + 1$ ⁠. A square is never negative, so $y \ge 1$ ⁠ for every $x$ ⁠: the whole parabola is above the $x$ ⁠-axis, with its vertex at $(-2, 1)$ ⁠. A negative discriminant says nothing about the sign of the roots; it says there are none.

Lesson 4 of 4 · About 25 minutes

From standard form to vertex form

Finding the vertex of y = ax² + bx + c, completing the square, and choosing a coefficient so that the parabola touches the axis at exactly one point.

Open the lesson

The 3 steps and their answers

Step 1 of 3
Open this step
Where the vertex is
Practice · Calculate
Question
What is the $x$ ⁠-coordinate of the vertex of $y = 2x^2 - 8x + 3$ ⁠?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
a
2fixed
b
-8fixed
c
3fixed
y = x² / The whole curve
Up or downdefaulthidden
Other settings
Mark vertex/nodes/graph/props/markVertex = false
Other settings
Mark axis/nodes/graph/props/markAxis = false
Accepted within
± 0.01
Computed from the Scene at this step’s setting
X of the vertex; NaN when a = 0/nodes/graph/outputs/vertexX
Wrong answers it recognises
- -2 ± 0.01
  Misconception: Uses x = b/(2a) for the vertex, dropping the minus sign.
  Feedback from the catalogue
  The $x$ ⁠-coordinate of the vertex is $x = -\dfrac{b}{2a}$ ⁠, with a minus. When you complete the square, $x^2 + bx$ ⁠ turns into $\left(x + \tfrac{b}{2}\right)^2$ ⁠, which is zero at $x = -\tfrac{b}{2}$ ⁠.
- 4 ± 0.01
  Misconception: Uses x = −b/a for the vertex, forgetting the 2 in the denominator.
  Feedback from the catalogue
  The vertex lies halfway between the roots, so $x = -\dfrac{b}{2a}$ ⁠: the two roots add up to $-\dfrac{b}{a}$ ⁠, and halfway is half of that.
Feedback for a right answer
$x = -\dfrac{-8}{2 \cdot 2} = \dfrac{8}{4} = 2$ ⁠. The axis of symmetry is $x = 2$ ⁠, and the vertex is $(2, 2 \cdot 4 - 16 + 3) = (2, -5)$ ⁠.

Hint
$x = -\dfrac{b}{2a}$ ⁠, with $a = 2$ ⁠ and $b = -8$ ⁠.

Worked solution
1. Identify the coefficients: $a = 2$ ⁠ and $b = -8$ ⁠.
2. Substitute into the formula: $x = -\dfrac{b}{2a} = -\dfrac{-8}{2 \cdot 2} = \dfrac{8}{4} = 2$ ⁠.
Two traps: the minus in the formula and the minus of $b$ ⁠ cancel, so the answer is positive ( $-2$ ⁠ forgets one of them); and the denominator is $2a = 4$ ⁠, not $a$ ⁠ ( $4$ ⁠ forgets the $2$ ⁠). Check by completing the square: $2x^2 - 8x + 3 = 2(x-2)^2 - 5$ ⁠, and the vertex is $(2, -5)$ ⁠.
Step 2 of 3
Open this step
Completing the square
Practice · Choose
Question
How do you write $y = x^2 + 6x + 5$ ⁠ in vertex form?

Correct answer
- Correct $y = (x+3)^2 - 4$ ⁠
Setup, choices and feedback
The Scene at this step
a
2
b
-8
c
3
y = x² / The whole curve
y = x²
Choices
Shown in a random order.
- Correct $y = (x+3)^2 - 4$ ⁠
- Wrong $y = (x-3)^2 - 4$ ⁠
  Misconception: Uses x = b/(2a) for the vertex, dropping the minus sign.
  Feedback from the catalogue
  The $x$ ⁠-coordinate of the vertex is $x = -\dfrac{b}{2a}$ ⁠, with a minus. When you complete the square, $x^2 + bx$ ⁠ turns into $\left(x + \tfrac{b}{2}\right)^2$ ⁠, which is zero at $x = -\tfrac{b}{2}$ ⁠.
- Wrong $y = (x+3)^2 + 5$ ⁠
  Misconception: Completes the square without subtracting the added square, so the constant term is unchanged.
  Feedback from the catalogue
  $(x+3)^2 = x^2 + 6x + 9$ ⁠, so the brackets added a $9$ ⁠ that was not in the original expression. To keep the function the same, you have to take it away again: $x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$ ⁠.
- Wrong $y = (x+6)^2 - 31$ ⁠
  Feedback
  What goes in the brackets is half of $b$ ⁠: $(x+6)^2 = x^2 + 12x + 36$ ⁠, not $x^2 + 6x$ ⁠.
Feedback for a right answer
Half of $6$ ⁠ is $3$ ⁠: $(x+3)^2 = x^2 + 6x + 9$ ⁠. The brackets added $9$ ⁠, so you take it away again: $y = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$ ⁠. The vertex is $(-3, -4)$ ⁠.

Worked solution
1. Take half the coefficient of $x$ ⁠: half of $6$ ⁠ is $3$ ⁠.
2. Expand the square: $(x+3)^2 = x^2 + 6x + 9$ ⁠. It gives the $x^2 + 6x$ ⁠, plus an extra $9$ ⁠.
3. Take away what was added: $x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$ ⁠.
Check: $(x+3)^2 - 4 = x^2 + 6x + 9 - 4 = x^2 + 6x + 5$ ⁠. The vertex is $(-3, -4)$ ⁠, which $x = -\dfrac{b}{2a} = -3$ ⁠ confirms.
Step 3 of 3
Open this step
Touch the axis once
Practice · Reach the target
Question
Right now the parabola $y = x^2 + bx + 4$ ⁠ does not touch the $x$ ⁠-axis. Choose a value for b so that it touches the axis at exactly one point, then press Check.

Goal
- 0 ± 0.001
  At the start of the step: -16
Setup, choices and feedback
The Scene at this step
a
1fixed
b
0
c
4fixed
y = x² / The whole curve
Up or down
Other settings
Mark roots/nodes/graph/props/markRoots = true
Read from the Scene at
B² − 4ac/nodes/graph/outputs/discriminant
One setting that reaches it
b
-4
Feedback for a right answer
$b^2 = 16$ ⁠, so $b = 4$ ⁠ or $b = -4$ ⁠: both answers are right. With $b = 4$ ⁠ it is $y = (x+2)^2$ ⁠, with its vertex at $(-2, 0)$ ⁠; with $b = -4$ ⁠ it is $y = (x-2)^2$ ⁠, with its vertex at $(2, 0)$ ⁠.

Feedback for a wrong answer
Not yet. Look at the vertex: it has to sit exactly on the $x$ ⁠-axis, neither above it nor below it.

Hint
Touching at one point means $\Delta = 0$ ⁠. Here $\Delta = b^2 - 4 \cdot 1 \cdot 4 = b^2 - 16$ ⁠.

Worked solution
Touching the $x$ ⁠-axis at one point means exactly one root, that is $\Delta = 0$ ⁠.
1. Here $a = 1$ ⁠ and $c = 4$ ⁠, so $\Delta = b^2 - 4 \cdot 1 \cdot 4 = b^2 - 16$ ⁠.
2. $b^2 - 16 = 0$ ⁠ gives $b^2 = 16$ ⁠, so $b = 4$ ⁠ or $b = -4$ ⁠.
Both answers are right: with $b = 4$ ⁠ the function is $(x+2)^2$ ⁠, which touches the axis at $x = -2$ ⁠; with $b = -4$ ⁠ it is $(x-2)^2$ ⁠, which touches it at $x = 2$ ⁠. Set b to either value.

Across the unit · 9 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

The shape and sign of a

The basic parabola

Through zero

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Up or down

Question

Correct answer

The Scene at this step

Choices

Worked solution

Narrow or wide

Question

Correct answer

The Scene at this step

Choices

Worked solution

Shifting a parabola

Three steps from the basic parabola

Which way?

Question

Correct answer

The Scene at this step

Choices

Worked solution

Hit the ring

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Read the vertex

Question

Correct answer

The Scene at this step

Choices

Worked solution

Roots and the discriminant

Where the graph meets the axis

Predict before you look

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

Work out the discriminant

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

When the discriminant is negative

Question

Correct answer

The Scene at this step

Choices

Worked solution

From standard form to vertex form

Where the vertex is

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Completing the square

Question

Correct answer

The Scene at this step

Choices

Worked solution

Touch the axis once