How to Attack Any Question
Before you reach for a formula, slow down for ten seconds and figure out what the question is actually asking and what it already handed you. Almost every mistake on a test is answering a slightly different question than the one on the page. Here's the move that fixes that.
🧭 The 5-Step Game Plan
Run this on every problem — graphing, solving, word problems, all of it.
🔎 What Is It Actually Asking?
The wording is a code. Here's what each phrase on your study guide is really asking for, and where to go for it.
🧱 Which Form Am I Looking At? Read Off What's Free
Recognizing the form is half the battle — each one hands you something for nothing before you do any work.
🛠️ Picking a Solving Tool
When the goal is "solve for x," first make it read = 0, then pick the lightest tool that fits.
▶️ The Game Plan in Action
Let's run all five steps on a real projectile problem from the study guide.
A ball is launched from a 60 m platform: h(t) = −5t² + 20t + 60.
(a) What's the maximum height? (b) When does it hit the ground?
① Find: two things — (a) a max height, (b) a landing time.
② Given: standard form, with a = −5. Since a < 0 it opens down, so a maximum really exists. ✓
③ Tool: "max height" = the vertex's y · "hits the ground" = set h(t) = 0 and solve.
④ Do it (a): t = −202(−5) = 2 sec. h(2) = −5(2)² + 20(2) + 60 = −20 + 40 + 60 = 80 m.
④ Do it (b): −5t² + 20t + 60 = 0 → divide by −5 → t² − 4t − 12 = 0 → (t − 6)(t + 2) = 0 → t = 6 or t = −2 → time can't be negative → t = 6 sec.
⑤ Check: a < 0 so a max makes sense ✓ · it starts at c = 60 m (the platform) ✓ · it lands (t = 6) after the peak (t = 2) ✓.
Reading a Quadratic
1 Recognizing a Quadratic
A quadratic is any equation whose highest power of x is 2 — that little ² is the giveaway. No higher exponent is allowed (no x³, no x⁴).
That x² term is what bends the graph into a smooth U-shaped curve called a parabola.
2 Standard Form
Every quadratic can be written this way. The three numbers each have a job:
3 Shape of the Graph
The sign of a decides which way the parabola opens:
Graphing the Parabola
4 Vertex Formula
The vertex is the turning point — the very bottom (or top) of the curve. Find its x-coordinate first:
Then plug that x back into the equation to get the matching y. Let's do it with our example y = 2x² + 4x + 5:
Here a = 2 and b = 4.
x = −42 × 2 = −44 = −1
y = 2(−1)² + 4(−1) + 5 = 2 − 4 + 5 = 3
vertex = (−1, 3)
5 Maximum vs. Minimum
The vertex is either the lowest or the highest point — and the sign of a tells you which:
In our example a = 2 (positive), so the vertex (−1, 3) is a minimum.
6 Axis of Symmetry
A parabola is a perfect mirror image of itself. The mirror line is a vertical line straight through the vertex:
Same formula as the vertex's x-value — that's no accident. For our example the axis of symmetry is x = −1. Whatever the curve does on one side, it mirrors on the other.
7 Y-Intercept
To find where the graph crosses the y-axis, set x = 0. Every x term disappears and you're left with c.
y = 2(0)² + 4(0) + 5 = 5 → y-intercept is (0, 5)
8 X-Intercepts
The x-intercepts are where the graph crosses the x-axis. On the x-axis the height is zero, so you set y = 0 and solve for x.
These are also called the roots or zeros. The next section is all about how to solve them.
Solving Quadratics
9 Factoring
Factoring rewrites the quadratic as two things multiplied together. The trick: find two numbers that multiply to c and add to b.
x² + x − 156 = 0
Need two numbers that multiply to −156 and add to 1.
→ 13 and −12 (13 × −12 = −156, 13 + (−12) = 1)
(x + 13)(x − 12) = 0
10 Zero Product Property
Here's why factoring works. If two things multiply to zero, at least one of them must be zero:
So set each factor equal to zero and solve each one:
(x + 13)(x − 12) = 0
x + 13 = 0 → x = −13
x − 12 = 0 → x = 12
11 Quadratic Formula
When a quadratic won't factor nicely, this formula always works. Just plug in a, b, and c:
The ± means you do it twice — once with plus, once with minus — which gives the two solutions.
Solve x² − 4x + 1 = 0 (try to factor it — you can't find two whole numbers, so reach for the formula).
Here a = 1, b = −4, c = 1.
First the discriminant: b² − 4ac = (−4)² − 4(1)(1) = 16 − 4 = 12 (positive → 2 real solutions)
x = −(−4) ± √122(1) = 4 ± √122
√12 ≈ 3.46, so x ≈ 3.73 or x ≈ 0.27
12 Discriminant
The part under the square root has a name: the discriminant. Its sign tells you how many real solutions you'll get — before you even finish solving.
(crosses x-axis twice)
(just touches x-axis)
(never touches x-axis)
Word Problems
13 Rectangle / Path Problems
When a border or path of width x wraps all the way around a shape, each dimension grows by x on both sides — so it grows by 2x total.
14 Volume Problems
Volume of a box multiplies its three dimensions. When a dimension is an expression with x, the volume becomes a quadratic (or cubic).
15 Projectile Problems
An object thrown or launched follows a parabola. Its height over time is modeled by:
Changing Forms
16 Completing the Square
This is the move that converts standard form into vertex form — handy when you want the vertex but the equation isn't factored.
y = x² − 8x − 4
y = (x − 4)² − 20
Now the vertex is easy to read straight off: (4, −20).
17 The Three Main Forms
Same parabola, three outfits. Each form hands you a different piece of information for free — so pick the form that matches what you need.
Graph y = 2x² + 4x + 5
We already found the key features above. Here are five points — pick x-values around the vertex and compute y. Tap a row to highlight it on the graph.
| x | y = 2x² + 4x + 5 | point |
|---|