Problem 19 - College Algebra with Corequisite Support

Question

For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. $$ f(x)=\frac{1}{2} x^{2}+3 x+1 $$

Step-by-Step Solution

Answer

The minimum value of the function is \(-\frac{7}{2}\), with the axis of symmetry at \(x = -3\).

Step 1: Identify the Quadratic Function Form

The quadratic function is given as \( f(x) = \frac{1}{2}x^2 + 3x + 1 \), which is in the form \( ax^2 + bx + c \) where \( a = \frac{1}{2} \), \( b = 3 \), and \( c = 1 \).

Step 2: Determine if Minimum or Maximum Value Exists

For a quadratic function \( ax^2 + bx + c \) in standard form, the graph is a parabola. The parabola opens upwards when \( a > 0 \) and downwards when \( a < 0 \). Here, \( a = \frac{1}{2} > 0 \), so the parabola opens upwards, which means there is a minimum value.

Step 3: Find the Axis of Symmetry

The axis of symmetry for a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Substitute the values \( b = 3 \) and \( a = \frac{1}{2} \) into the formula:\[x = -\frac{3}{2 \times \frac{1}{2}} = -\frac{3}{1} = -3\]Thus, the axis of symmetry is \( x = -3 \).

Step 4: Calculate the Minimum Value

To find the minimum value of the quadratic function when \( x = -3 \), substitute \( x = -3 \) back into \( f(x) \):\[f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + 1 \= \frac{1}{2} \times 9 - 9 + 1 \= \frac{9}{2} - 9 + 1 \= \frac{9}{2} - \frac{18}{2} + \frac{2}{2} \= -\frac{7}{2}\]Therefore, the minimum value is \(-\frac{7}{2}\).

Key Concepts

Axis of Symmetry

The axis of symmetry is a critical component in understanding the geometry of a parabola formed by a quadratic function. It is a vertical line that divides the parabola into two mirror-image halves. This line runs through the vertex of the parabola, which is either the highest or lowest point, depending on the orientation of the parabola. For any quadratic function in the form of \( ax^2 + bx + c \), the axis of symmetry can be found using the formula:

• \( x = -\frac{b}{2a} \)

In this example, the function is \( f(x) = \frac{1}{2}x^2 + 3x + 1 \). Here, \( a = \frac{1}{2} \) and \( b = 3 \). By substituting these values into the formula, you calculate:

• \( x = -\frac{3}{2 \times \frac{1}{2}} = -3 \)

So, the axis of symmetry is the line \( x = -3 \). This line is extremely useful because it not only helps in graphing the parabola but also simplifies finding the vertex of the parabola.

Minimum Value

Quadratic functions can have either a minimum or maximum value, depending on the direction in which the parabola opens. The quadratic function \( f(x) = \frac{1}{2}x^2 + 3x + 1 \) has values that form a parabola opening upwards.
This upward opening indicates that the vertex of the parabola represents its lowest point, or the minimum value.

• Because the coefficient \( a = \frac{1}{2} \) is positive, the parabola opens upwards.

To find the minimum value, substitute the x-coordinate of the vertex into the quadratic function. We already found that the axis of symmetry is \( x = -3 \), which means the vertex (and minimum point) is at \( x = -3 \). Substitute \( x = -3 \) into the function:

• \( f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + 1 \)
• \( = \frac{9}{2} - 9 + 1 \)
• \( = -\frac{7}{2} \)

The minimum value of the function is \( -\frac{7}{2} \), indicating the vertex point is \((-3, -\frac{7}{2})\). This value tells us the lowest point reached by \( f(x) \) when graphing this quadratic function.

Parabola Orientation

The orientation of a parabola is determined by the sign of the leading coefficient, \( a \), in the quadratic function \( ax^2 + bx + c \). In this exercise, the quadratic function is \( f(x) = \frac{1}{2}x^2 + 3x + 1 \), where the coefficient \( a \) is \( \frac{1}{2} \). Since \( a > 0 \), the parabola is oriented upwards.Understanding parabola orientation is vital:

• If \( a > 0 \), the parabola opens upwards, resembling a "U" shape. It implies the vertex will have the function's minimum value.
• If \( a < 0 \), it opens downwards, similar to an upside-down "U", indicating the vertex has the maximum value.

This orientation not only affects the placement of the vertex but also the general behavior and range of the quadratic function's values.
An upward orientation, as in this case, tells us that all values of the function will be greater than or equal to the minimum value, \( -\frac{7}{2} \). Thus, as \( x \) moves away from the axis of symmetry both to the left and right, the function values will increase.