The parabola and its five landmarks — vertex, axis of symmetry, y-intercept and zeros — read from a graph, a table or an equation; standard, factored and vertex form and what each one reveals. Then solving: square roots, factoring and the zero-product rule, completing the square, and the quadratic formula with the discriminant that predicts two, one or no real roots. A thrown ball modeled by h = −16t² + 32t + 48 turns every feature into a sentence: where it starts, how high it goes, when it lands. Study cards, hints, a practice quiz at three levels, and four workshop activities that check themselves.
A parabola. The first thing to notice is that it is a mirror — the left half matches the right.
Five landmarks: the vertex, the axis of symmetry, the y-intercept, and the two zeros. Each form of the equation hands you a different one of them for free — which is why there are three forms.
What people get wrong
⚠️People often think…
x² = 36, so x = 6.
That is half an answer. x = ±6, because (−6)² is also 36. Every square root step in a quadratic produces two signs, and the parabola's own symmetry is the reason — it crosses twice.
A mirror crosses twice. Write both signs.
⚠️People often think…
y = (x + 4)² − 5 has its vertex at (4, −5).
Vertex form is y = a(x − h)² + k, so the sign inside flips: x + 4 means h = −4. The vertex is (−4, −5). The k outside does not flip — only the number inside the parentheses does.
Inside flips, outside doesn't.
⚠️People often think…
(4 ± 2√3) ÷ 2 is 4 ± √3.
The 2 on the bottom divides the WHOLE top, not just the root part. It is 2 ± √3. Dividing only half the numerator is the single most expensive slip in the quadratic formula.
Every term on top gets divided.
⚠️People often think…
In a height equation, c is the maximum height.
c is where it STARTED, at t = 0 — the height of the hand that threw it. The peak is the vertex, so find t = −b ÷ 2a first, then put that t back in to get the height.
“How high did it start” and “how high did it get” are different questions.
Worth knowing cold
Three forms, and what each one gives you free
They all describe the same curve. Which one you want depends on which landmark the question is asking for.
Standardy = ax² + bx + c. Hands you c, the y-intercept, on sight.
Factoredy = a(x − p)(x − q). Hands you the zeros: p and q.
Vertexy = a(x − h)² + k. Hands you the vertex (h, k) — mind the flip.
Axisx = −b ÷ 2a. Always straight through the vertex.
a > 0Opens up. The vertex is the lowest point — a minimum.
a < 0Opens down. The vertex is the highest point — a maximum.
Two signs, every time
1Watch one
Solve x² = 36.
Take the square root of both sides.
6 squared is 36 — but so is −6 squared.
The parabola crosses twice, one on each side.
x = ±6. Both, always.
2Do one with me
Find the vertex of y = (x + 4)² − 5.
Vertex form is a(x − h)², so the sign inside flips. h =
The number outside does not flip. k =
So the vertex is
💬One sentence, then you move on
Why does every square root step in a quadratic give two answers?
3Try one
Solve x² = 5x. Write both answers.
I want a hint first
Do not divide by x — x might be zero, and dividing by zero throws that answer away. Move the 5x over and factor.
💬Last one — then you're done here
Why does dividing both sides by x lose a solution?
Where this goes
Where this lives
Anything thrown or dropped follows a parabola: a free throw, a punt, water from a hose. The vertex is the top of the arc and the zeros are where it leaves and lands.
What this feeds
That closes Algebra I. Geometry picks it up next, starting with proof and congruence.
Name one thing you have watched follow a curve like this.
One card at a time — tap “Show me” to check yourself, then Next. Start at Foundation; when those feel easy, climb.
Helpful Hints
🧮 Solving any quadratic — the routine
1 · Get 0 on one side, in standard form: x² + 4x = 12 becomes x² + 4x − 12 = 0. 2 · No x-term? Square roots, both signs. 3 · Does it factor? (x + 6)(x − 2) = 0, so x = −6 or 2. 4 · If not, the formula: write a, b, c WITH their signs, compute b² − 4ac first, then x = (−b ± √disc) / 2a. 5 · Check one root in the ORIGINAL equation: (−6)² + 4(−6) = 36 − 24 = 12 ✓.
Zero, roots, factor, formula, check.
⛰️ The whole unit in one idea — one curve, three forms
Every quadratic is one parabola wearing three outfits. Standard form y = x² − 2x − 8 shows where it starts on the y-axis (−8). Factored form y = (x − 4)(x + 2) shows where it crosses the x-axis (4 and −2). Vertex form y = (x − 1)² − 9 shows its turning point (1, −9). Solving the equation means finding the zeros; graphing means finding the vertex and the intercepts; both are the same curve seen from a different side.
x² = 36 so x = 6. Half an answer: x = ±6. Every square root in a quadratic has two signs.
Reading y = (x + 4)² − 5 as vertex (4, −5). Inside the parentheses the sign flips: h = −4, vertex (−4, −5).
Using b = 3 in the formula when the equation says −3x. Copy a, b, c with their signs before you start.
Dividing the root by 2a but not the −b: (4 ± 2√3)/2 is 2 ± √3, not 4 ± √3.
Solving x² = 5x by dividing by x. That loses x = 0. Move it over: x(x − 5) = 0.
Calling c the maximum height. c is the START (t = 0). The peak is the vertex — find t = −b/2a first.
📋 Four methods and the discriminant — the chart
Situation
Method
Worked example
No x-term
square roots, ±
x² − 9 = 0 → x² = 9 → x = ±3
Factors nicely
factor, zero product
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2, 3
a = 1, b even
complete the square
x² + 6x + 5 = 0 → (x + 3)² = 4 → x = −1, −5
Anything at all
quadratic formula
x² − 3x − 4 = 0 → x = (3 ± 5)/2 = 4, −1
b² − 4ac > 0
two real roots
x² + 4x + 1: 16 − 4 = 12
b² − 4ac = 0
one double root (touches)
x² − 6x + 9: 36 − 36 = 0
b² − 4ac < 0
no real roots (misses)
x² + 2x + 5: 4 − 20 = −16
🎯 How the test will ask
“Solve” — an equation, maybe not yet equal to 0. Set it to 0, pick a method, give BOTH roots.
“Find the vertex / axis of symmetry / y-intercept” of y = ax² + bx + c. x = −b/2a, plug in, x = 0 for c.
“Rewrite in vertex form” or “in factored form.” Complete the square, or factor; then say what the form shows.
“How many real solutions?” — the discriminant b² − 4ac, no solving needed.
A graph — “which equation matches?” Read the vertex, the direction, the y-intercept, and test one point.
A ball, a rocket, a diver: h = −16t² + v₀t + h₀. Landing = zero, peak = vertex, start = h(0). Reject the negative time.
A rectangle with a given area — write w(w + 3) = 40, solve, and throw out the negative width.
✅ Before the test, can you…
Name a, b and c with their signs, and say which way the parabola opens?
Find the vertex, axis of symmetry, y-intercept and zeros of y = x² − 4x + 3 and sketch it from those five points?
Solve x² + 6x + 5 = 0 by factoring, by completing the square, and by the formula, and get the same two roots?
Write the quadratic formula from memory and simplify (4 ± √12)/2 to 2 ± √3?
Use the discriminant to say two, one or zero real roots without solving?
Convert y = x² − 8x + 3 to vertex form and read the vertex from it?
Take h = −16t² + 32t + 48 and find when the ball lands (3 s), how high it goes (64 ft), and why t = −1 is thrown out?
Pick your level
Look back at anything you missed — the hint that appeared is exactly what to reread tonight.
How sure did you feel?
Workshop
Work like an algebraist: name the five landmarks of a parabola, put completing the square in order, sort twelve equations by how many real roots they have, then settle a claim about a thrown ball with the vertex and the zeros. Every activity checks itself, and hints are free.
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