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Chapter 1: Problem 56

Make a table of values and sketch the graph of the equation. Find the \(x\) - and \(y\) -intercepts and test for symmetry. $$y=x^{2}+2$$

Short Answer

Expert verified
The graph is a parabola opening upwards with a y-intercept at (0, 2) and no x-intercepts; it is symmetric about the y-axis.

Step by step solution

01

Create a Table of Values

To create a table of values, choose several values for \( x \), substitute them into the equation \( y = x^2 + 2 \), and solve for \( y \). For example, you might choose \( x = -2, -1, 0, 1, 2 \). Calculate \( y \) for each of these values. The table looks like this: \(\begin{array}{c|c} x & y \\hline -2 & 6 \ -1 & 3 \ 0 & 2 \ 1 & 3 \ 2 & 6 \\end{array}\)
02

Sketch the Graph

Using the table of values from Step 1, plot the points on a graph. For instance, for \( x = -2 \), plot the point \(( -2, 6 )\). Continue this for each \( x \) value from the table. Connect the points with a smooth curve to show the parabolic shape of the equation \( y = x^2 + 2 \).
03

Find the x-intercepts

To find the \( x \)-intercepts, set \( y = 0 \) in the equation and solve for \( x \). Set \( 0 = x^2 + 2 \), which simplifies to \( x^2 = -2 \). As there are no real solutions to this equation (since \( x^2 \) cannot be negative in real numbers), there are no \( x \)-intercepts.
04

Find the y-intercept

To find the \( y \)-intercept, set \( x = 0 \) in the equation \( y = x^2 + 2 \). This yields \( y = 0^2 + 2 = 2 \). Therefore, the \( y \)-intercept is \( (0, 2) \).
05

Test for Symmetry

Test for symmetry with respect to the \( y \)-axis by substituting \( -x \) in place of \( x \) in the equation and seeing if you obtain an equivalent expression. Substitute into \( y = (-x)^2 + 2 \), which simplifies back to \( y = x^2 + 2 \). Since the equation is equivalent, it is symmetric with respect to the \( y \)-axis. The graph is not symmetric with respect to the \( x \)-axis or the origin, as substituting \( -y \) or \( (-x, -y) \) does not satisfy the original equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola
A parabola is a U-shaped curve that represents the graph of a quadratic function. In the equation \( y = x^2 + 2 \), the term \( x^2 \) signifies that it is indeed a quadratic function.
  • Parabolas can open upwards or downwards based on the degree and sign of the quadratic term. Here, the coefficient of \( x^2 \) is positive, so it opens upwards.
  • The vertex of a parabola is the highest or lowest point on the graph, depending on whether it opens upward or downward.
For this equation, the vertex is not translated left or right (since there is no \( x \)-term), but it is shifted upward by 2 units, making the minimum point at \( (0, 2) \). This feature is deduced from the equation \( y = x^2 + 2 \), indicating a vertical shift from \( y = x^2 \).
x-intercepts
The \( x \)-intercepts are the points where the graph crosses or touches the \( x \)-axis. To find these intercepts, you need to set\( y = 0 \) in the equation and solve for \( x \). For the equation \( y = x^2 + 2 \), substituting \( 0 \) for \( y \), gives you \( x^2 + 2 = 0 \). Simplifying this equation results in \( x^2 = -2 \), which has no real solutions since the square of a real number cannot be negative.
  • This absence of \( x \)-intercepts means that the parabola does not cross the \( x \)-axis.
  • It stays entirely above the axis because of its vertex being 2 units above the origin and opening upwards.
y-intercept
The \( y \)-intercept is the point where the graph intersects the \( y \)-axis. To find this, set \( x = 0 \) in the equation \( y = x^2 + 2 \). Doing this calculation, you get \( y = 0^2 + 2 \), which simplifies to \( y = 2 \). Thus, the \( y \)-intercept is the point \( (0, 2) \).
  • This is a critical point for sketching the graph, as it shows where the parabola begins its journey upwards from the \( y \)-axis.
  • Combined with the directional sense of the curve (upwards), this aids in accurately plotting and understanding the function's graph.
Symmetry in Graphs
A graph can display various types of symmetry, but for parabolas, we often look for symmetry around the axes. For the equation \( y = x^2 + 2 \), symmetry with respect to the \( y \)-axis is indicated by re-evaluating the function with \( -x \): Substitute \( -x \) into the equation, as in \( y = (-x)^2 + 2 \). This simplifies back to \( y = x^2 + 2 \), showing that the graph is indeed symmetric with respect to the \( y \)-axis.
  • This symmetry implies that the left and right sides of the graph are mirror images of each other.
  • It helps predict function values without recalculating when moving equidistantly from the center to either direction on the \( x \)-axis.
Scan for symmetry about the \( x \)-axis or the origin, you'll find that it's not applicable here as those transformations do not equate back to the original expression.
Table of Values
Constructing a table of values is essential for graphing and understanding quadratic functions. The table lets us observe the output \( y \) values for chosen inputs \( x \). For the function \( y = x^2 + 2 \):
  • Select several \( x \)-values. Common choices are integers around zero, such as -2, -1, 0, 1, and 2, to view the function's behavior on both sides of the \( y \)-axis.
  • Substitute each \( x \)-value into the equation to find the corresponding \( y \)-value, forming pairs of points \((x, y)\).
The table of values facilitates constructing the graph by marking these critical points. After plotting, join these points in a smooth curve to visualize the parabola clearly.

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Most popular questions from this chapter

State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$\frac{x}{x+y}=\frac{1}{1+y}$$

A city has streets that run north and south, and avenues that run east and west, all equally spaced. Streets and avenues are numbered sequentially, as shown in the figure. The walking distance between points \(A\) and \(B\) is 7 blocks - that is, 3 blocks cast and 4 blocks north. To find the straight-line distances \(d\), we must use the Distance Formula. (a) Find the straight-line distance (in blocks) between \(A\) and \(B\) (b) Find the walking distance and the straight-line distance between the corner of 4 th St. and 2 nd Ave. and the corner of 11 th St. and 26 th Ave. (c) What must be true about the points \(P\) and \(Q\) if the walking distance between \(P\) and \(Q\) equals the straightline distance between \(P\) and \(Q\) ?

Factor the expression completely. $$\left(1+\frac{1}{x}\right)^{2}-\left(1-\frac{1}{x}\right)^{2}$$

(a) Find the radius of each circle in the pair, and the distance between their centers; then use this information to determine whether the circles intersect. (i) \((x-2)^{2}+(y-1)^{2}=9\) \((x-6)^{2}+(y-4)^{2}=16\) (ii) \(x^{2}+(y-2)^{2}=4\) \((x-5)^{2}+(y-14)^{2}=9\) (iii) \((x-3)^{2}+(y+1)^{2}=1\) \((x-2)^{2}+(y-2)^{2}=25\) (b) How can you tell, just by knowing the radii of two circles and the distance between their centers, whether the circles intersect? Write a short paragraph describing how you would decide this and draw graphs to illustrate your answer.

The rational expression $$\frac{x^{2}-9}{x-3}$$ is not defined for \(x=3 .\) Complete the tables and determine what value the expression approaches as \(x\) gets closer and closer to \(3 .\) Why is this reasonable? Factor the numerator of the expression and simplify to see why. $$\begin{array}{c|c} \hline x & \frac{x^{2}-9}{x-3} \\ \hline 2.80 & \\ 2.90 & \\ 2.95 & \\ 2.99 & \\ 2.999 & \end{array}$$ $$\begin{array}{|l|l|} \hline x & \frac{x^{2}-9}{x-3} \\ \hline 3.20 & \\ 3.10 & \\ 3.05 & \\ 3.01 & \\ 3.001 & \\ \hline \end{array}$$
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