Problem 45 - Algebra 2

Question

Determine whether each function has a maximum or a minimum value and find the maximum or minimum value. Then state the domain and range of the function. $$ f(x)=2 x+2 x^{2}+5 $$

Step-by-Step Solution

Answer

The function has a minimum value of 4.5. Its domain is \(( -\infty, \infty )\) and the range is \([4.5, \infty)\).

Step 1: Identify the Type of Function

The function given is \( f(x) = 2x + 2x^2 + 5 \). Notice that it is a quadratic function because the highest power of \( x \) is 2. Quadratic functions are generally of the form \( ax^2 + bx + c \). In this case, \( a = 2 \), \( b = 2 \), and \( c = 5 \). Since \( a \) is positive, the parabola opens upwards, so the function has a minimum value.

Step 2: Find the Vertex Formula

The vertex of a parabola given by \( ax^2 + bx + c \) can be found using the formula for the \( x \)-coordinate of the vertex, \( x = -\frac{b}{2a} \).

Step 3: Calculate the x-coordinate of the Vertex

Substitute \( b = 2 \) and \( a = 2 \) into the vertex formula: \( x = -\frac{2}{2 \times 2} = -\frac{1}{2} \). This is the x-coordinate where the minimum value occurs.

Step 4: Calculate the Minimum Value

Substitute \( x = -\frac{1}{2} \) back into the function to find the minimum value: \[f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) + 2\left(-\frac{1}{2}\right)^2 + 5 \ = -1 + 2\left(\frac{1}{4}\right) + 5 = -1 + \frac{1}{2} + 5 = -0.5 + 5 = 4.5\]Thus, the minimum value of the function is 4.5.

Step 5: State the Domain of the Function

The domain of any quadratic function is all real numbers \(( -\infty, \infty )\) because you can substitute any real number into \( x \) and compute \( f(x) \).

Step 6: State the Range of the Function

Since the parabola opens upwards and has a minimum value of 4.5, the range of the function is \([4.5, \infty)\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point that provides valuable information about the graph. It represents the peak point if the parabola opens downwards or the lowest point if it opens upwards. For the function \( f(x) = 2x + 2x^2 + 5 \), which is a quadratic function, the parabola opens upwards because the coefficient \( a = 2 \) is positive.
To find the vertex, we use the vertex formula for the x-coordinate: \( x = -\frac{b}{2a} \). Here, \( b = 2 \) and \( a = 2 \), so the x-coordinate of the vertex is \(-\frac{1}{2}\).
To find the corresponding y-coordinate of the vertex (the minimum value), substitute the x-value back into the function. This calculation gives you the vertex at \((-\frac{1}{2}, 4.5)\).
This vertex tells us that the lowest point of the parabola is at this coordinate, and since the parabola opens upwards, every other point will be above this point.

Domain and Range

Understanding the domain and range of a function helps you grasp what values are possible. The domain refers to all possible x-values (inputs) that can be used in the function, while the range refers to the y-values (outputs) that the function can produce.
For quadratic functions, like \( f(x) = 2x + 2x^2 + 5 \), the domain is always all real numbers. This is because you can plug any real number into \( x \) and still get a valid output. Thus, the domain is expressed as \(( -\infty, \infty )\).
The range is a bit different. Since our parabola has a minimum value at 4.5 and it opens upwards, this means that y-values start from 4.5 and go to positive infinity. Therefore, the range is expressed as \([4.5, \infty)\).

• Domain: \(( -\infty, \infty )\)
• Range: \([4.5, \infty)\)

Minimum and Maximum Values

Quadratic functions can have either a minimum or maximum value depending on the orientation of the parabola. This is dictated by the coefficient \( a \) in the standard form \( ax^2 + bx + c \).
If \( a > 0 \), the parabola opens upwards and the function has a minimum value at the vertex. For our function \( f(x) = 2x + 2x^2 + 5 \), \( a = 2 \), so the parabola opens upwards, ensuring a minimum value.
We found the minimum value by substituting \( x = -\frac{1}{2} \) back into the original function, resulting in a minimum value of 4.5.
Since the parabola doesn't open downwards (because \( a \) is not negative), there's no maximum value for this function. The concept holds that the minimum marks the lowest y-value the function can achieve, and beyond the vertex, y-values continue to increase towards infinity.