SAT Quadratics in Desmos: Roots, Vertices and the Requested Value

Enter:

y = x^2 - 8x + 12

An x-window from 0 to 8 and a y-window from −6 to 14 will display both zeros and the vertex. Select the curve to show points of interest. The x-intercepts are (2,0) and (6,0); the vertex is (4,−4).

x² − 8x + 12 = (x − 2)(x − 6). Setting the product to zero gives x = 2 or x = 6. Their sum is 8 and their product is 12. A prompt asking for the larger solution therefore requires 6, not 8 or the vertex’s x-coordinate 4.

x² − 8x + 12 = (x − 4)² − 4. Since the square is nonnegative, the minimum is −4 at x = 4. For a restricted domain such as x ≥ 5, that unrestricted vertex is excluded; the minimum on x ≥ 5 is instead f(5) = −3. Always apply the given domain.

To solve x² − 8x + 12 = 5, graph y = x² − 8x + 12 and y = 5. Their intersection x-values solve that equation. Algebra gives (x − 4)² − 4 = 5, so (x − 4)² = 9 and x = 1 or 7. Using the original quadratic’s zeros 2 and 6 would solve a different equation.

For y = x² − 10x + 21, factoring gives (x − 3)(x − 7), so zeros are 3 and 7. Completing the square gives (x − 5)² − 4, so the vertex is (5,−4). Graph those results, then explain why the minimum y-value is −4 while the larger zero is 7.