Worksheet for graphing quadratic functions in vertex form, with space to sketch graphs and answer questions about vertex, domain, range, and intercepts.

Let’s solve each problem step by step.

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Problem 1: y = -x²

This is a quadratic in vertex form. The general vertex form is:

> y = a(x - h)² + k

In this case, we can rewrite y = -x² as:

> y = -1(x - 0)² + 0

So, the vertex (h, k) is (0, 0).

- Since the coefficient of x² is negative (-1), the parabola opens down.
- Domain of any quadratic is all real numbers → (-∞, ∞)
- Range: Since it opens down and vertex is at y=0, the highest point is 0. So range is y ≤ 0, or (-∞, 0]
- Axis of symmetry: For vertex form, it’s always x = h → here, x = 0
- x-intercept: Set y = 0 → 0 = -x² → x² = 0 → x = 0 → so (0, 0)
- y-intercept: Set x = 0 → y = -(0)² = 0 → so (0, 0)

✔ All checks out.

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Problem 2: y = (x - 3)²

Again, vertex form: y = a(x - h)² + k

Here, a = 1, h = 3, k = 0 → vertex is (3, 0)

- Coefficient of x² is positive (1) → opens up
- Domain: All real numbers → (-∞, ∞)
- Range: Opens up, vertex at y=0 → lowest point is 0 → y ≥ 0, or [0, ∞)
- Axis of symmetry: x = h → x = 3
- x-intercept: Set y = 0 → 0 = (x - 3)² → x - 3 = 0 → x = 3 → (3, 0)
- y-intercept: Set x = 0 → y = (0 - 3)² = 9 → (0, 9)

✔ All checks out.

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Final Answer:

Problem 1: y = -x²

- Vertex: (0, 0)
- Opens Up or Down: Down
- Domain: (-∞, ∞)
- Range: (-∞, 0]
- Axis of Symmetry: x = 0
- x-intercept: (0, 0)
- y-intercept: (0, 0)

Problem 2: y = (x - 3)²

- Vertex: (3, 0)
- Opens Up or Down: Up
- Domain: (-∞, ∞)
- Range: [0, ∞)
- Axis of Symmetry: x = 3
- x-intercept: (3, 0)
- y-intercept: (0, 9)