Problem 17 - College Algebra

Question

For the following exercises, determine where is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. $$f(x)=4 x^{2}+x-1$$

Step-by-Step Solution

Answer

Axis of symmetry: \( x = -\frac{1}{8} \); Minimum value: \(-\frac{17}{16}\)

Step 1: Identify the quadratic function

The given function is a quadratic function in the form of \( ax^2 + bx + c \). In this exercise, \( a = 4 \), \( b = 1 \), and \( c = -1 \).

Step 2: Determine the axis of symmetry

The axis of symmetry for a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). Substituting the values of \( b \) and \( a \), we get:\[ x = -\frac{1}{2 \times 4} = -\frac{1}{8} \]Thus, the axis of symmetry is \( x = -\frac{1}{8} \).

Step 3: Determine the direction of the parabola

Since the coefficient \( a \) is positive (\( a = 4 \)), the parabola opens upwards, indicating that the function has a minimum value.

Step 4: Find the minimum value

To find the minimum value of the function, substitute \( x = -\frac{1}{8} \) into the function \( f(x) = 4x^2 + x - 1 \) and calculate:\[f\left(-\frac{1}{8}\right) = 4\left(-\frac{1}{8}\right)^2 + \left(-\frac{1}{8}\right) - 1\]\[= 4 \times \frac{1}{64} - \frac{1}{8} - 1\]\[= \frac{1}{16} - \frac{1}{8} - 1\]\[= \frac{1}{16} - \frac{2}{16} - \frac{16}{16}\]\[= -\frac{17}{16}\]Thus, the minimum value is \(-\frac{17}{16}\).

Key Concepts

Axis of Symmetry

In the world of quadratic functions, the axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. Think of it as the spine of the parabola. For any quadratic function expressed as \( ax^2 + bx + c \), you can find the axis of symmetry using the formula \( x = -\frac{b}{2a} \).
When you use this formula, you're determining the x-coordinate of the vertex of the parabola.
It's crucial because the axis of symmetry gives you valuable insight into the function's behavior.
Using our example \( f(x) = 4x^2 + x - 1 \), we identified that \( a = 4 \) and \( b = 1 \).
Plugging these into the formula:

• \( x = -\frac{1}{2 \times 4} \)
• \( x = -\frac{1}{8} \)

Therefore, the axis of symmetry is \( x = -\frac{1}{8} \), telling us where the parabola will be mirrored on both sides.

Minimum Value

Quadratic functions can have either a minimum or a maximum value depending on the direction in which their parabola opens.
If you're looking at a graph of the function and the directions seem puzzling, finding the minimum or maximum helps you get a clear picture of the most extreme value the function reaches.
In our quadratic function \( f(x) = 4x^2 + x - 1 \), since the parabola opens up (which we'll discuss shortly), it has a minimum value.
You find this minimum at the vertex, the point where the parabola changes direction.
To calculate it, after deriving the axis \( x = -\frac{1}{8} \), substitute back into the function:

• Calculate \( f\left(-\frac{1}{8}\right) \)
• After simplification, you find \( f(x) = -\frac{17}{16} \)

So, the minimum value is \(-\frac{17}{16}\), representing the lowest point of the parabola on the coordinate plane.

Parabola Direction

The direction of the parabola is a key feature that helps determine the nature of the function's extreme value—either a minimum or maximum.
To know whether a quadratic function opens upwards or downwards, look no further than the coefficient \( a \) in \( ax^2 + bx + c \).
Here are the rules that guide us:

• If \( a > 0 \), the parabola opens upwards; it resembles a smile \((\cup)\) and indicates the function has a minimum value.
• If \( a < 0 \), the parabola opens downwards; it resembles a frown \((\cap)\) and indicates the function has a maximum value.

For the function \( f(x) = 4x^2 + x - 1 \), the coefficient \( a = 4 \) is positive.
Thus, the parabola opens upwards, confirming that the function indeed has a minimum value.
The direction influences the entire graph's shape and behavior, making it a crucial element to understand when analyzing quadratic functions.