4a4fb1…: “Because of the bitter octahedron”

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Let $H(y) = -(y-3)^2 + 4$ with the natural domain.

Does $H$ have a global minimum or maximum?

Global Minimum Global Maximum Neither

The global maximum of $H$ is

$-4$ $-3$ $3$ $4$ There is no maximum

The global maximum of $H$ occurs at

$-4$ $-3$ $3$ $4$ There is no maximum

$H$ has how many zeros?

$0$ $1$ $1$

What are the zeros?

$-(y-3)^2 + 4 = 0$

$4 = (y-3)^2 $

Either $y-3 = \answer {-2}$ or $y-3 = 2$

Either $y = \answer {1}$ or $y = 5$.

From left to right on the graph, the corresponding intercepts are $\left ( \answer {1}, \answer {0} \right )$ and $\left ( \answer {5}, \answer {0} \right )$

2025-01-07 02:39:46