Problem 6 - Precalculus Mathematics for Calculus

Question

The graph of a quadratic function \(f\) is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of \(f\). (c) Find the domain and range of \(f\). $$f(x)=-\frac{1}{2} x^{2}-2 x+6$$ (GRAPH CAN'T COPY)

Step-by-Step Solution

Answer

The vertex is (-2, 8), the maximum value is 8, domain is all real numbers, and range is \((-\infty, 8]\).

Step 1: Identify the Form of the Quadratic Function

The function given is \( f(x) = -\frac{1}{2}x^2 - 2x + 6 \), which is in the standard quadratic form \( ax^2 + bx + c \), where \( a = -\frac{1}{2} \), \( b = -2 \), and \( c = 6 \).

Step 2: Find the Vertex of the Quadratic

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex \((h, k)\) can be found using the formula \( h = -\frac{b}{2a} \). Here, \( b = -2 \) and \( a = -\frac{1}{2} \). Thus, \( h = -\frac{-2}{2(-\frac{1}{2})} = -2 \). To find \( k \), substitute \( x = -2 \) into the function: \[ f(-2) = -\frac{1}{2}(-2)^2 - 2(-2) + 6 = -2 + 4 + 6 = 8 \]. Therefore, the vertex is \((-2, 8)\).

Step 3: Determine the Maximum or Minimum Value

Since the coefficient of \( x^2 \), \( a = -\frac{1}{2} \), is negative, the parabola opens downwards, indicating that the vertex is a maximum. Therefore, the maximum value of \( f \) is \( 8 \) at \( x = -2 \).

Step 4: Identify the Domain of the Function

The domain of any quadratic function \( ax^2 + bx + c \) is all real numbers because there are no restrictions on the values that \( x \) can take. Thus, the domain is \( (-\infty, \infty) \).

Step 5: Determine the Range of the Function

Since the parabola opens downwards, the range is limited from the maximum value at the vertex downwards. Therefore, the range of \( f(x) \) is \( (-\infty, 8] \).

Key Concepts

Vertex of a Parabola

In the world of quadratic functions, understanding the vertex of a parabola is crucial. The vertex represents the highest or lowest point on a parabola, depending on its orientation. To find the vertex of a parabola represented by the quadratic function in standard form, \( f(x) = ax^2 + bx + c \), we use the formula \( h = -\frac{b}{2a} \) to find the x-coordinate.

• "\(a\)" determines the parabola's direction (open upwards if positive, open downwards if negative).
• "\(b\)" helps shift the parabola left or right.

Once you find \( h \), substitute this back into the original function to calculate \( k \), the y-coordinate of the vertex. This gives us the coordinates \((h, k)\) of the vertex. For the given function, \( f(x) = -\frac{1}{2}x^2 - 2x + 6 \), the vertex is \((-2, 8)\). This is not only a coordinate but also tells us critical information about the function's shape.

Maximum and Minimum of a Function

Quadratic functions like the one given exhibit a crucial feature: they either have a maximum or a minimum value but not both. The vertex of the parabola is key in determining this value. Since the parabola's orientation dictates whether it is a maximum or minimum:

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

For the function \( f(x) = -\frac{1}{2}x^2 - 2x + 6 \), the parabola opens downwards because \( a = -\frac{1}{2} \) is negative. Hence, this function has a maximum value at its vertex. This maximum value is the y-coordinate of the vertex, which is \( 8 \). Therefore, the maximum value of \( f(x) \) is \( 8 \), achieved when \( x = -2 \). Understanding the maximum or minimum helps in interpreting the function's behavior.

Domain and Range

Understanding the domain and range of quadratic functions enables us to see their entire spectrum of behavior. The **domain** refers to all possible values of \( x \) that you can plug into the function without breaking any mathematical rules. For any standard quadratic function \( ax^2 + bx + c \), the domain is all real numbers because quadratic equations can accommodate any real x-value.

• Domain of the quadratic function: \((-\infty, \infty)\)

The **range**, however, is the set of all possible output values (\( f(x) \)). It is dependent on whether the parabola opens up or down:

• If the parabola opens upwards, the range starts from the vertex's y-value and goes to positive infinity, \( [k, \infty) \).
• If it opens downwards, like in our function, the range extends from negative infinity to the vertex's y-value, \( (-\infty, 8] \).

For our specific function \( f(x) = -\frac{1}{2}x^2 - 2x + 6 \), because it opens downwards, the range is \( (-\infty, 8] \), bounded above by the vertex.