Problem 19
Question
Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$f(x)-(x-1)^{2}+2$$
Step-by-Step Solution
Verified Answer
The axis of symmetry of the function \(f(x) = (x-1)^2 + 2\) is x = 1. The domain of the function is all real numbers and it's range is \([2, +\infty)\)
1Step 1: Identify the Vertex
Since the function is given in vertex form, \(f(x) = (x - h)^2 + k\), the vertex of the parabola is at point (h, k). For the given function \(f(x) = (x - 1)^2 + 2\), the vertex is at point (1, 2).
2Step 2: Determine the Axis of Symmetry
The axis of symmetry for a parabola in vertex form is x = h. For the given function, the axis of symmetry is x = 1.
3Step 3: Graph the Function and Find Intercepts
Plot the vertex and draw the axis of symmetry. Then to find the x-intercept, set \(f(x) = 0\), and solve for x. In this case, the equation is \((x - 1)^2 + 2 = 0\). The solutions for x will give the x-intercepts. As for the y intercept, plug in 0 into the function for x and solve for y.
4Step 4: Determine the Direction of the Parabola
In the given function, the coefficient of \(x^2\) is positive, thus the parabola opens upward.
5Step 5: Find the Domain and Range
For a quadratic function, the domain is all real numbers. The range is determined by the vertex and the direction of the parabola. As this parabola opens upward and the y-coordinate of the vertex is 2, the range is \([2, +\infty)\)
Key Concepts
Vertex FormAxis of SymmetryDomain and RangeParabola
Vertex Form
The vertex form of a quadratic function is a handy way to describe and manipulate parabolas. It's written as:
Here, \((h, k)\) represents the vertex of the parabola,
For the function \( f(x) = (x - 1)^2 + 2 \), the vertex is at point \((1, 2)\). This point provides valuable information about the location of the parabola on the coordinate plane.
- \( f(x) = a(x - h)^2 + k \)
Here, \((h, k)\) represents the vertex of the parabola,
- \(a\) is a coefficient that affects the width and direction of the parabola.
For the function \( f(x) = (x - 1)^2 + 2 \), the vertex is at point \((1, 2)\). This point provides valuable information about the location of the parabola on the coordinate plane.
Axis of Symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. This line passes through the vertex of the parabola. Because of its symmetry, any point on the parabola has a corresponding point located directly across the axis.
For quadratic functions in vertex form \( f(x) = (x - h)^2 + k \), the axis of symmetry can be easily found as \( x = h \). This is because the vertex is \((h, k)\), and the parabola is symmetrical about this point.
In our example, \( f(x) = (x - 1)^2 + 2 \), the axis of symmetry is \( x = 1 \). This means every point on one side of the parabola is mirrored on the other side along the line \( x = 1 \).
For quadratic functions in vertex form \( f(x) = (x - h)^2 + k \), the axis of symmetry can be easily found as \( x = h \). This is because the vertex is \((h, k)\), and the parabola is symmetrical about this point.
In our example, \( f(x) = (x - 1)^2 + 2 \), the axis of symmetry is \( x = 1 \). This means every point on one side of the parabola is mirrored on the other side along the line \( x = 1 \).
Domain and Range
When studying quadratic functions, it's important to understand the concepts of domain and range. The domain of a function refers to the set of all possible input values (x-values) that make the function work. For any quadratic function, including parabolas, the domain is always all real numbers because you can plug any real number into the quadratic equation to solve for \( f(x) \).
The range, on the other hand, is the set of all possible output values (y-values) from the function. For our upward-opening parabola \( f(x) = (x - 1)^2 + 2 \), the lowest point is the vertex at \((1, 2)\). This means the y-values start at 2 and go upwards, resulting in a range of \([2, +\infty)\).
The range, on the other hand, is the set of all possible output values (y-values) from the function. For our upward-opening parabola \( f(x) = (x - 1)^2 + 2 \), the lowest point is the vertex at \((1, 2)\). This means the y-values start at 2 and go upwards, resulting in a range of \([2, +\infty)\).
Parabola
A parabola is a symmetrical, U-shaped curve that is the graph of a quadratic function. Understanding its properties is crucial when dealing with quadratic functions.
The function \( f(x) = (x - 1)^2 + 2 \) describes a parabola that opens upwards due to a positive \(a = 1\). It has a vertex at \((1, 2)\), making this point the minimum point of the curve.
- The direction in which a parabola opens (upward or downward) is determined by the coefficient \(a\) in the vertex form equation. If \(a\) is positive, the parabola opens upward; if negative, it opens downward.
- The vertex is a key feature of a parabola, representing either the minimum or maximum point, depending on the orientation.
- The axis of symmetry helps divide the parabola into two equal halves, making it easier to graph and analyze.
- Intercepts are the points where the parabola crosses the x and y axes.
The function \( f(x) = (x - 1)^2 + 2 \) describes a parabola that opens upwards due to a positive \(a = 1\). It has a vertex at \((1, 2)\), making this point the minimum point of the curve.
Other exercises in this chapter
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