Algebra 2A — Functions & Quadratics

The vertex — a parabola's turning point

continues from lesson 2 — values defined earlier in the course stay live here

ELI5: every parabola has a single lowest (or highest) point, the vertex, sitting at x = −b/2a — exactly halfway between the two roots. For y = x² − 6x + 5 that's x = 3, the bottom of the U.

q_{v}\left(x\right) = x^{2} - 6 \cdot x + 5

a parabola opening upward

vertex_{x} = \frac{6}{2}

3

x = −b/2a = 6/2 = 3

✓ pass

q_{v}(vertex_{x}) <= q_{v}(vertex_{x} - 0.5) and q_{v}(vertex_{x}) <= q_{v}(vertex_{x} + 0.5)

the vertex sits below both neighbours — it is the minimum

Try it yourself: find the x of the vertex of x² − 8x + 10. (It's the lowest point of the U.)

q_{w}\left(x\right) = x^{2} - 8 \cdot x + 10

vx_{you} = 4

✏️ Your turn: find the x at the bottom of x² − 8x + 10. The check confirms it sits below both neighbours — so it never reveals the answer.

✓ pass

q_{w}(vx_{you}) <= q_{w}(vx_{you} - 0.5) and q_{w}(vx_{you}) <= q_{w}(vx_{you} + 0.5)

green when your x is the true minimum

Where next: Algebra 2B — exponentials, the logarithms that undo them, and the sums of sequences.