Problem

Classify the extremum of \(f(x)=2(x-1)^{2}-3\). Use the leading coefficient to justify the opening direction, state the vertex, extremum type, value and location, and write the range.

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What this problem is really about

Minimum versus maximum is determined by the leading coefficient’s sign, not by whether the vertex output is positive or negative. We’ll use that sign for the opening, read the vertex, classify its output as the extremum, and translate the attained boundary into interval notation for the range.

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The vertex is (−3, 1), so the minimum is 1 at x = −3. The h and k values are reversed.

The vertex is found from the standard coefficients without using 2(x − 1)² − 3, producing a different extremum and range.

Because the leading coefficient is positive, the parabola opens up. Its vertex is (1, −3), so it has a minimum value of −3 at x = 1 and range [−3, ∞).

The parabola opens up with vertex (1, −3), but it has a maximum because k is negative. The opening, not the sign of k, determines minimum versus maximum.

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Classify the extremum of \(f(x)=2(x-1)^{2}-3\). Use the leading coefficient to justify the opening direction, state the vertex, extremum type, value and location, and write the range.

What this problem is really about

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