Problem 33
Question
Graph the parabola whose equation is given $$y=x^{2}+2 x+1$$
Step-by-Step Solution
Verified Answer
The graph of the parabola given by the equation \(y = x^2 + 2x + 1\) has its vertex at the point (-1,0) and is symmetrical about the line \(x = -1\). The parabola opens upwards.
1Step 1: Identify the standard form of the parabola
The standard form of a parabola is \(y = a(x-h)^{2} + k\), where (h, k) is the vertex of the parabola. The given equation \(y = x^{2} + 2x + 1\) can also be written as \(y = (x+1)^{2}\).
2Step 2: Identify the vertex of the parabola
The vertex in this case is given by (-1, 0). When the equation is written as \(y = a(x-h)^{2} + k\), the vertex is given by the values \(-h\) and \(-k\). Here, h and k both equal -1 so the vertex is (-1, 0).
3Step 3: Identify the axis of symmetry
For a parabola of the form \(y = a(x-h)^{2} + k\), the axis of symmetry is given by \(x = h\). In this case, the axis of symmetry is \(x = -1\).
4Step 4: Plot the Parabola
Begin by plotting the vertex, and the axis of symmetry as a dotted vertical line. Since the coefficient of \(x^{2}\) is positive, the parabola opens upwards. Find a few more values to the left and right of the vertex for a more detailed graph. By substituting \(x = -2, 0\), we get corresponding \(y\) values as \(1, 1\). By symmetry, copy these points to the right of the axis of symmetry. Join these points to complete the graph
Key Concepts
Vertex of a ParabolaAxis of SymmetryStandard Form of a Parabola
Vertex of a Parabola
The vertex of a parabola is a major conceptual point which represents the highest or lowest point on the graph, depending on whether it opens upwards or downwards, respectively. In relation to the equation of the parabola in standard form, which is expressed as
\[y = a(x-h)^{2} + k\]
the vertex is represented by the coordinate pair \(h, k\). It's important to understand that \(h\) and \(k\) are derived directly from the equation, where \(h\) is the x-coordinate and \(k\) is the y-coordinate of the vertex.
For the given exercise, with the parabolic equation \(y=x^{2}+2x+1\), completing the square transforms it into the equivalent \(y=(x+1)^{2}\). From this form, the vertex can be determined as \(h=-1\) and \(k=0\), thus the vertex is at \( (-1, 0) \). The reason we take \(h=-1\) instead of \(h=1\) from \(x+1\) is because the standard form shows \(x-h\), so the sign is switched.
In graphing the parabola, finding and plotting the vertex is the first step and serves as an anchor point from which the graph is shaped.
\[y = a(x-h)^{2} + k\]
the vertex is represented by the coordinate pair \(h, k\). It's important to understand that \(h\) and \(k\) are derived directly from the equation, where \(h\) is the x-coordinate and \(k\) is the y-coordinate of the vertex.
For the given exercise, with the parabolic equation \(y=x^{2}+2x+1\), completing the square transforms it into the equivalent \(y=(x+1)^{2}\). From this form, the vertex can be determined as \(h=-1\) and \(k=0\), thus the vertex is at \( (-1, 0) \). The reason we take \(h=-1\) instead of \(h=1\) from \(x+1\) is because the standard form shows \(x-h\), so the sign is switched.
In graphing the parabola, finding and plotting the vertex is the first step and serves as an anchor point from which the graph is shaped.
Axis of Symmetry
The axis of symmetry is a vertical line that splits the parabola into two mirrored halves. For any parabola described by the standard form, \[y = a(x-h)^{2} + k\], the axis of symmetry is always the line \(x = h\). This is because the vertex, which lies on this line, is the point where the parabola turns or 'bends'.
Given our example, since our vertex is \( (-1, 0) \), the equation of our axis of symmetry is \(x = -1\). If you draw a vertical line through \(x = -1\) on a graph, you'd have the axis of symmetry for this parabola. It's useful because once you plot points on one side of the axis of symmetry, you can 'reflect' them to the other side to easily complete the parabolic shape when graphing by hand.
Understanding the axis of symmetry is crucial for graphing the parabola and for solving problems involving symmetry, such as finding the optimal path or maximizing an area under a curve.
Given our example, since our vertex is \( (-1, 0) \), the equation of our axis of symmetry is \(x = -1\). If you draw a vertical line through \(x = -1\) on a graph, you'd have the axis of symmetry for this parabola. It's useful because once you plot points on one side of the axis of symmetry, you can 'reflect' them to the other side to easily complete the parabolic shape when graphing by hand.
Understanding the axis of symmetry is crucial for graphing the parabola and for solving problems involving symmetry, such as finding the optimal path or maximizing an area under a curve.
Standard Form of a Parabola
The standard form of a parabola is perhaps one of the most recognizable and important algebraic forms:
\[y = a(x-h)^{2} + k\].
Here, the variables \(a\), \(h\), and \(k\) are essential elements. The coefficient \(a\) affects the direction and 'width' of the parabola. If \(a > 0\), it opens upwards; if \(a < 0\), it opens downwards. The larger the absolute value of \(a\), the 'steeper' or 'narrower' the parabola is. The variables \(h\) and \(k\) determine the location of the vertex, as explained earlier.
When working with the problem \(y=x^{2}+2x+1\), we are using this standard form as our starting point for graphing. We place the vertex and use the axis of symmetry, then plot additional points and use the mirroring property of the parabola relative to the axis of symmetry to fill in the rest of the graph. Knowing how to write and interpret the parabola in standard form is an essential skill for algebra and beyond, as it simplifies the process of graphing and analyzing parabolas in various mathematical contexts.
\[y = a(x-h)^{2} + k\].
Here, the variables \(a\), \(h\), and \(k\) are essential elements. The coefficient \(a\) affects the direction and 'width' of the parabola. If \(a > 0\), it opens upwards; if \(a < 0\), it opens downwards. The larger the absolute value of \(a\), the 'steeper' or 'narrower' the parabola is. The variables \(h\) and \(k\) determine the location of the vertex, as explained earlier.
When working with the problem \(y=x^{2}+2x+1\), we are using this standard form as our starting point for graphing. We place the vertex and use the axis of symmetry, then plot additional points and use the mirroring property of the parabola relative to the axis of symmetry to fill in the rest of the graph. Knowing how to write and interpret the parabola in standard form is an essential skill for algebra and beyond, as it simplifies the process of graphing and analyzing parabolas in various mathematical contexts.
Other exercises in this chapter
Problem 32
Solve each quadratic equation using the quadratic formula. $$x^{2}=4 x-7$$
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Express each function as a set of ordered pairs. $$f(x)=2 x+3 ; \text { domain: }[-1,0,1]$$
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Solve each equation by the method of your choice. Simplify irrational solutions, if possible. $$(3 x-4)^{2}=16$$
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Solve each quadratic equation by first factoring the perfect square trinomial on the left side. Then apply the square root property. Simplify radicals, if possi
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