i Your toolkit
Everything in this studio comes back to one idea: the vertex is the answer.
y = a(x − h)² + k
Vertex (h, k). Watch the sign: (x + 4) means h = −4.
Vertex (h, k). Watch the sign: (x + 4) means h = −4.
a > 0 → opens up (minimum). a < 0 → opens down (maximum). Larger |a| → narrower.
The vertex is found at x = −b / (2a) in standard form, then substitute to get the y-value.
Turning point = maximum height, maximum profit, maximum area, best price.
x-intercepts (roots) = hitting the ground, break-even points, where the jet lands.
Given the vertex and one other point, substitute to find a.
1 Choose your questions
Click any tile to open that problem. Solve them in whatever order you like — each one is marked as you go.
Solved 0 / 10 · First-try score 0 / 10
Question
Question 1 of 10
🔑 The big ideas
Vertex form is the key. y = a(x − h)² + k shows the vertex (h, k) instantly — and in every real-world model the vertex is the maximum or minimum of something.
Watch the sign of h. y = a(x + 4)² + k has vertex (−4, k), not (4, k). This catches people out on arch and jet problems.
Projectile motion. Height h(t) = −5t² + vt + c. The vertex gives the maximum height and the time it happens; the roots give launch and landing times.
Optimisation. Area, profit and revenue problems always reduce to a quadratic. Find the vertex — its x-value is the best choice, its y-value is the best result.
Setting up the model. Name the unknown, write a constraint (like a fixed perimeter or a fixed number of tickets), then build the quadratic and complete the square.
Finding a. If you know the vertex (h, k) and one other point, substitute that point and solve for a. This is how you build an equation from a real arch, cable or jet.
Roots = break-even. Whenever the model crosses zero — ground level, zero profit — you are solving a(x − h)² + k = 0.