Problem 42
Question
Find the vertex and axis of symmetry of the associated parabola for each quadratic function.Then find at least two additional points on the parabola and sketch the parabola by hand. $$f(x)=x^{2}-6 x+1$$
Step-by-Step Solution
Verified Answer
The vertex of the parabola is (3, -8) and the axis of symmetry is the line \(x = 3\). Additional points on the parabola include (2, -5) and (4, -11).
1Step 1: Find the Vertex
For the quadratic function \(f(x) = x^{2} - 6x + 1\), we have \(a = 1\) and \(b = -6\). The x-coordinate of the vertex is given by \(- \frac{b}{2a} = - \frac{-6}{2*1} = 3\). Substituting \(x = 3\) into the function, the y-coordinate of the vertex is \(f(3) = (3)^{2} - 6*3 + 1 = -8\). So, the vertex of the parabola is at the point \((3, -8)\).
2Step 2: Find the Axis of Symmetry
The equation of the axis of symmetry can be obtained directly from the x-coordinate of the vertex. It is the vertical line \(x = 3\).
3Step 3: Find Additional Points
To graph the parabola, we need at least two more points. We can choose any x-values we want, but smaller and bigger values than the x-coordinate of the vertex would give us a better sketch of the parabola. Let's pick \(x = 2\) and \(x = 4\). For \(x = 2\), the y-value is \(f(2) = (2)^{2} - 6*2 + 1 = -5\), so we get the point \((2, -5)\). For \(x = 4\), the y-value is \(f(4) = (4)^{2} - 6*4 + 1 = -11\), giving us the point \((4, -11)\).
4Step 4: Sketch the Parabola
Plot the vertex, the axis of symmetry, and the additional points on the graph. Draw a smooth curve through these points to represent the parabola.
Key Concepts
Vertex of a ParabolaAxis of SymmetryGraphing Parabolas
Vertex of a Parabola
Understanding the vertex of a parabola is vital in graphing and analyzing quadratic functions. Simply put, the vertex is the highest or lowest point on the parabola, depending on its direction (upward or downward). In the given exercise, we identify the vertex for the function
Once we have the x-coordinate, substituting it back into the original function gives us the corresponding y-coordinate. So, for our function with
The vertex serves as a crucial reference point for graphing the entire parabola and also for understanding its properties, such as the maximum or minimum value of the quadratic function.
f(x) = x^2 - 6x + 1 by using the formula x = -b / (2a). This represents the horizontal coordinate of the vertex.Once we have the x-coordinate, substituting it back into the original function gives us the corresponding y-coordinate. So, for our function with
a = 1 and b = -6, the vertex's x-coordinate is 3. Substituting x = 3 into the function, we find the y-coordinate to be -8. Hence, the vertex of our parabola is at (3, -8).The vertex serves as a crucial reference point for graphing the entire parabola and also for understanding its properties, such as the maximum or minimum value of the quadratic function.
Axis of Symmetry
The axis of symmetry is a fundamental characteristic of parabolas. It's a straight line that vertically divides the parabola into two mirror images. In other words, it's the line where if you folded the parabola along it, both sides would align perfectly. For a quadratic function of the form
In the exercise we're analyzing, the axis of symmetry for the parabola associated with
f(x) = ax^2 + bx + c, the axis of symmetry always passes through the vertex, and its equation is x = -b / (2a).In the exercise we're analyzing, the axis of symmetry for the parabola associated with
f(x) = x^2 - 6x + 1 is the vertical line x = 3. This line not only helps in graphing but also in finding other points on the parabola. By selecting x-values that are equidistant from the axis of symmetry, we can find y-values that are equal, thus demonstrating the parabola's symmetric nature.Graphing Parabolas
Once you have found the vertex and axis of symmetry, you're ready to start graphing a parabola. In the question provided, the next step is to find additional points to get a better sketch of the shape of the parabola. By choosing x-values on either side of the vertex, and finding the corresponding y-values, you can sketch a more accurate curve.
For instance, using
For instance, using
x = 2 and x = 4 as in the exercise's solution, we obtain the points (2, -5) and (4, -11) respectively. Plotting these alongside the vertex and the axis of symmetry will give a clear outline of the parabola. Then, gently sketch a smooth curve through the vertex, making sure the curve is symmetrical about the axis of symmetry. With practice, graphing parabolas becomes an intuitive process, especially when you understand the significance of each component in its composition.Other exercises in this chapter
Problem 41
Solve the quadratic equation by using the quadratic formula. Find only real solutions. $$x^{2}+2 x-1=0$$
View solution Problem 41
Decide if each function is odd, even, or neither by using the definitions. $$f(x)=x^{5}-2 x$$
View solution Problem 42
The height of a ball that is thrown directly upward from a point 200 feet above the ground with an initial velocity of 40 feet per second is given by \(h(t)=-16
View solution Problem 42
Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{2 x-1}+2=x$$
View solution