Problem 20
Question
19-28 \(=\) A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$ f(x)=x+x^{2} $$
Step-by-Step Solution
Verified Answer
Standard form is \( x^2 + x + 0 \), vertex at \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \), minimum value is \(-\frac{1}{4}\).
1Step 1: Write the given function
The given quadratic function is \( f(x) = x + x^2 \). We need to express it in standard form, which is \( f(x) = ax^2 + bx + c \).
2Step 2: Rearrange function to standard form
Rearranging the given quadratic function, we can write: \( f(x) = x^2 + x \). So, \( a = 1 \), \( b = 1 \), and \( c = 0 \). Now it is in standard form \( f(x) = x^2 + x + 0 \).
3Step 3: Sketch the graph
The quadratic \( f(x) = x^2 + x \) has no constant term, shifts from the origin due to the \( b \) term. It opens upwards since \( a > 0 \). The vertex can be used to help plot, but further analysis is in the next step.
4Step 4: Find the vertex (h, k)
The vertex form of a quadratic is \( (h, k) \) from \( x = \frac{-b}{2a} \). Calculate \( x = \frac{-1}{2(1)} = -\frac{1}{2} \). Substitute \(-\frac{1}{2}\) into \( f(x) \) to find \( y \): \[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]. Vertex is \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \).
5Step 5: Identify maximum or minimum value
Since \( a = 1 > 0 \), the parabola smiles (opens upwards) and thus has a minimum. The minimum value is the \( y \)-coordinate of the vertex: \( -\frac{1}{4} \).
Key Concepts
Standard Form of Quadratic FunctionGraphing Quadratic FunctionsVertex of a Parabola
Standard Form of Quadratic Function
A quadratic function is generally expressed in a specific format known as the **standard form**. This form is represented as \( f(x) = ax^2 + bx + c \), where:
To transform a function into its standard form, you simply rearrange its terms so that the \( x^2 \) term comes first, followed by the \( x \) term, and finally the constant term. For the function \( f(x) = x + x^2 \), rearrangement gives us \( f(x) = x^2 + x + 0 \), conveniently showing us the values \( a = 1 \), \( b = 1 \), and \( c = 0 \).
This format is essential because it provides the foundational elements to further analyze and graph the quadratic function.
- \( a \), \( b \), and \( c \) are constants.
- \( a \) is not equal to zero, because if \( a \) equals zero, the term \( ax^2 \) would disappear, and the function would no longer be quadratic.
To transform a function into its standard form, you simply rearrange its terms so that the \( x^2 \) term comes first, followed by the \( x \) term, and finally the constant term. For the function \( f(x) = x + x^2 \), rearrangement gives us \( f(x) = x^2 + x + 0 \), conveniently showing us the values \( a = 1 \), \( b = 1 \), and \( c = 0 \).
This format is essential because it provides the foundational elements to further analyze and graph the quadratic function.
Graphing Quadratic Functions
Graphing quadratic functions involves plotting their corresponding parabola based on the standard form equation, \( f(x) = ax^2 + bx + c \). A few key properties drive the graph's appearance:
For the quadratic function \( f(x) = x^2 + x \), this graph will be an upward-opening parabola because \( a = 1 \) is positive.
The axis of symmetry, busily slicing the graph, can be determined as \( x = -\frac{1}{2} \). With these details, a neatly symmetrical parabola can be plotted on a graph.
- The coefficient \( a \) determines the parabola’s direction: if \( a > 0 \), it opens upwards, like a smile, and if \( a < 0 \), it opens downwards, like a frown.
- The vertex is a crucial point on the parabola, standing at its maximum or minimum.
- The axis of symmetry, which can be found using the formula \( x = \frac{-b}{2a} \), splits the parabola into two symmetrical halves.
For the quadratic function \( f(x) = x^2 + x \), this graph will be an upward-opening parabola because \( a = 1 \) is positive.
The axis of symmetry, busily slicing the graph, can be determined as \( x = -\frac{1}{2} \). With these details, a neatly symmetrical parabola can be plotted on a graph.
Vertex of a Parabola
The vertex is a vital point of the parabola that gives deep insights into the quadratic function’s graph. It is the peak or the bottom of the curve, depending on how the parabola opens.
The vertex's x-coordinate can be calculated using the formula \( x = \frac{-b}{2a} \). Substituting the values for \( f(x) = x^2 + x \), we get:
So, the vertex of the parabola is at \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \).
This vertex represents the minimum point of the parabola since the graph opens upwards. Understanding the vertex’s importance helps in determining the shape and turning points of the quadratic graph.
The vertex's x-coordinate can be calculated using the formula \( x = \frac{-b}{2a} \). Substituting the values for \( f(x) = x^2 + x \), we get:
- \( x = \frac{-1}{2(1)} = -\frac{1}{2} \)
- \( f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \)
So, the vertex of the parabola is at \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \).
This vertex represents the minimum point of the parabola since the graph opens upwards. Understanding the vertex’s importance helps in determining the shape and turning points of the quadratic graph.
Other exercises in this chapter
Problem 19
Evaluate the function at the indicated values. $$ \begin{array}{l}{f(x)=2|x-1|} \\ {f(-2), f(0), f\left(\frac{1}{2}\right), f(2), f(x+1), f\left(x^{2}+2\right)}
View solution Problem 20
\(17-22=\) Use \(f(x)=3 x-5\) and \(g(x)=2-x^{2}\) to evaluate the expression. $$ \begin{array}{ll}{\text { (a) }(f \circ f)(-1)} & {\text { (b) }(g \circ g)(2)
View solution Problem 20
Sketch the graph of the function by first making a table of values. $$ f(x)=\frac{x}{|x|} $$
View solution Problem 20
If \(g(x)=x^{2}+4 x\) with \(x \geq-2,\) find \(g^{-1}(5)\)
View solution