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Graphing calculators are powerful tools that help us visualize mathematical functions, especially when dealing with complex equations like quadratics. By graphing the function on such a device, we can quickly identify important features of the graph. For quadratic functions, these features often include:

• The direction in which the parabola opens (upward or downward), determined by the sign of the coefficient of the square term, \(a\).
• The vertex of the parabola, which serves as the highest or lowest point on the graph, depending on the orientation of the parabola.

Using a graphing calculator allows us to obtain a visual representation of the function. This visual aid makes it easier to approximate values like the minimum or maximum, which is found at the vertex for quadratics. In this exercise, by graphing \(f(x) = x^2 + 1.79x - 3.21\), you would see a parabola opening upwards, indicating the presence of a minimum value at its vertex.

The vertex of a parabola is a pivotal point because it reveals either the minimum or maximum value of a quadratic function depending on the parabola's direction.

• An upward-opening parabola has its minimum value at the vertex.
• A downward-opening parabola has its maximum value at the vertex.

The vertex can be found algebraically using the formula \(-\frac{b}{2a}\), where \(a\) and \(b\) are coefficients from the standard quadratic expression \(ax^2 + bx + c\). For the given function \(f(x) = x^2 + 1.79x - 3.21\), substituting \(a = 1\) and \(b = 1.79\) into the vertex formula gives the x-coordinate of the vertex as \(-0.895\).
To find the corresponding y-value or the function's value at this point, substitute \(-0.895\) back into the function. This simple substitution yields the minimum value of the function at this vertex.

Quadratic functions typically have either a minimum or a maximum value at their vertex.

• If the coefficient \(a\) is positive, like in \(f(x) = x^2 + 1.79x - 3.21\), the function opens upwards, indicating a minimum at the vertex.
• If \(a\) were negative, the parabola would open downwards, indicating a maximum value at its vertex.

In this exercise, since \(a = 1\), the function has a minimum, which can be calculated exactly by substituting the x-coordinate of the vertex back into the function.
Thus, the minimum value calculated both graphically and algebraically (approximately \(-4.42\)) verifies the consistency and reliability of these methods for determining the extreme values of a quadratic function.