Question

Vertex Form: `y=a\left(x-h\right)^{2}+k`

The vertex is at`\left(h,\ k\right)`.

The axis of symmetry is `x=h`.

The minimum or maximum value is `k`.

Answer 1

That's correct. A few concise additions and reminders:

- General vertex form: y = a(x − h)^2 + k
- Vertex: (h, k)
- Axis of symmetry: x = h
- If a > 0 the parabola opens upward and k is the minimum value (range [k, ∞)).
- If a < 0 it opens downward and k is the maximum value (range (−∞, k]).

- Effect of parameters:
- a: vertical stretch/compression and reflection (|a| > 1 stretch, 0 < |a| < 1 compression, a < 0 reflects).
- h: horizontal shift (right by h if h > 0, left if h < 0).
- k: vertical shift (up by k if k > 0, down if k < 0).

- Converting from standard form y = ax^2 + bx + c to vertex form:
1. Compute h = −b/(2a).
2. Compute k = f(h) (plug h into the quadratic).
3. Then y = a(x − h)^2 + k.

- Completing-the-square method (same result):
Example: y = 2x^2 − 8x + 3
1. Factor a from x-terms: y = 2(x^2 − 4x) + 3
2. Complete square inside: x^2 − 4x + 4 = (x − 2)^2, so add/subtract 4 inside:
y = 2[(x^2 − 4x + 4) − 4] + 3 = 2(x − 2)^2 − 8 + 3
3. Simplify: y = 2(x − 2)^2 − 5
Vertex: (2, −5), axis x = 2, minimum −5 (since a = 2 > 0).

If you want, give me a quadratic and I’ll convert it to vertex form and identify the vertex, axis, and range.