Problem 5 - Beginning and Intermediate Algebra

Question

a) Does the function attain a minimum or maximum value at its vertex? b) Find the vertex of the graph of \(f(x)\) c) What is the minimum or maximum value of the function? d) Graph the function to verify parts a) \(-c\) ). $$f(x)=-\frac{1}{2} x^{2}+4 x-6$$

Step-by-Step Solution

Answer

a) The function attains a maximum value at its vertex because the leading coefficient, \(a\) is negative. b) The vertex is at the point \((4, 2)\). c) The maximum value of the function is \(2\). d) Upon graphing the function \(f(x) = -\frac{1}{2}x^2 + 4x - 6\), we can confirm that our answers for parts a), b), and c) are consistent with the graph.

Step 1: Identify Characteristics of the Quadratic Function

Quadratic functions are in the form of \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In our case, \(f(x) = -\frac{1}{2}x^2 + 4x - 6\), so \(a = -\frac{1}{2}\), \(b = 4\), and \(c = -6\). To determine whether the function has a minimum or maximum value at its vertex, we need to identify the sign of the coefficient \(a\). Since our coefficient \(a\) is negative, the function has a maximum value at its vertex.

Step 2: Find the Vertex of the Graph

The vertex of a quadratic function can be found using the formula: \(x_{vertex} = \frac{-b}{2a}\). Using the values of \(a = -\frac{1}{2}\) and \(b=4\), we can find the \(x\)-coordinate of the vertex. \[x_{vertex} = \frac{-4}{2(-\frac{1}{2})} = \frac{-4}{-1} = 4\] Next, we need to find the \(y\)-coordinate of the vertex by plugging the \(x\)-coordinate into our function: \[y_{vertex} = f(x_{vertex}) = f(4) = -\frac{1}{2}(4)^2 + 4(4) - 6 = -8 + 16 - 6 = 2\] The vertex of the graph is at the point \((4, 2)\).

Step 3: Find the Maximum Value of the Function

Since our function attains a maximum value at its vertex, the maximum value is the \(y\)-coordinate of the vertex. Therefore, the maximum value of the function is \(2\).

Step 4: Graph the Function to Verify Parts a) - c)

To verify our findings, we can graph the function and observe its vertex and the maximum value. Plotting the function \(f(x) = -\frac{1}{2}x^2 + 4x - 6\), we can see that it has a maximum value at \(y=2\) and its vertex is at the point \((4, 2)\), as determined earlier. This confirms our solution to parts a), b), and c).

Key Concepts

Vertex of a Quadratic

Understanding the vertex of a quadratic function is crucial to exploring its properties and behavior on a graph. The vertex represents the peak or the lowest point of a parabola, which is the shape formed by a quadratic equation. This vital point is found using the formula \( x_{vertex} = \frac{-b}{2a} \), where \( a \), \( b \), and \( c \) are coefficients from the standard quadratic form \( ax^2 + bx + c \). For the quadratic function \( f(x) = -\frac{1}{2}x^2 + 4x - 6 \), we identify \( a = -\frac{1}{2} \) and \( b = 4 \). Applying our formula gives us the \( x \)-coordinate of the vertex:\[ x_{vertex} = \frac{-4}{2(-\frac{1}{2})} = 4 \]Now, to find the \( y \)-coordinate, substitute \( x = 4 \) back into the function:\[ y_{vertex} = f(4) = -\frac{1}{2}(4)^2 + 4(4) - 6 = 2 \]Hence, the vertex of the graph is at the point \((4, 2)\). It's not only the turning point of the parabola but also the maximum or minimum value point of the function.

• Use the vertex formula \( x_{vertex} = \frac{-b}{2a} \).
• Plug \( x_{vertex} \) back in to find \( y_{vertex} \).

Maximum and Minimum Values

Determining whether a quadratic function has a maximum or minimum value is simple if you know the sign of the coefficient \( a \). The graph of a quadratic can either open upwards or downwards. This orientation tells you if the function achieves a maximum or minimum at its vertex.

• If \( a > 0 \), the parabola opens upwards, and the vertex is the minimum point.
• If \( a < 0 \), the parabola opens downwards, and the vertex is the maximum point.

For our function \( f(x) = -\frac{1}{2}x^2 + 4x - 6 \), the coefficient \( a = -\frac{1}{2} \) is less than zero, indicating the parabola opens downward. Therefore, the vertex \((4, 2)\) gives the maximum value of the function, which is \( y = 2 \). This property allows you to quickly determine the extremum of a quadratic function by merely inspecting \( a \).

Graphing Quadratic Equations

Graphing quadratic equations allows one to visualize the shape and key features of the function. The graph of a quadratic equation \( ax^2 + bx + c \) is a parabola, and graphing it can confirm the vertex, axis of symmetry, and whether it opens upward or downward. To graph a quadratic function like \( f(x) = -\frac{1}{2}x^2 + 4x - 6 \):

• Identify the vertex, which is \((4, 2)\) in our example.
• Note that the axis of symmetry is the vertical line \( x = 4 \).
• Plot additional points on either side of the vertex to ensure a smooth curve.
• Consider the direction of the parabola as dictated by the sign of \( a \). Since \( a \) is negative, the parabola opens downward.

By plotting the function, one can see the symmetry and extremum point visually. It's a useful tool to verify calculations and deepen understanding of quadratic behavior. Graphing calculators or software can provide precise plots, guiding students through a clear illustration of these concepts. Visualizing these functions enhances comprehension and confirms theoretical outcomes like maximum or minimum values.