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Chapter 3: Problem 2

True or False. The point (-2,-6) is on the graph of the equation \(x=2 y-2\).

Short Answer

Expert verified
False

Step by step solution

01

Understand the equation and point

First, recall that the given equation is in the form of a linear equation: \(x = 2y - 2\). And the point given is (-2, -6). We need to determine if this point satisfies the equation.
02

Substitute the coordinates into the equation

Substitute the coordinates of the point (-2, -6) into the equation: \(x = 2y - 2\)This means substitute \(x = -2\) and \(y = -6\).
03

Perform substitution

Replace \(x\) with -2 and \(y\) with -6 in the equation:\(-2 = 2(-6) - 2\)
04

Simplify the equation

Simplify the right-hand side of the equation:\(-2 = -12 - 2\)\(-2 = -14\)
05

Compare both sides

Observe that the left side, \(-2\), is not equal to the right side, \(-14\). Therefore, the initial equation is not satisfied with the point (-2, -6).
06

Conclusion

Since the left side does not equal the right side after substitution, the point (-2, -6) does not lie on the graph of the equation \(x = 2y - 2\). Therefore, the statement is False.

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Key Concepts

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

Point Substitution
Point substitution is a method used to determine if a particular point lies on the graph of a given equation. It involves replacing the variables in the equation with the coordinates of the point in question. Specifically, for an equation with variables x and y, you substitute the x-coordinate and the y-coordinate of the point into the respective variables in the equation. This method essentially checks if the ordered pair satisfies the equation. For instance, if we have the equation \(x = 2y - 2\) and the point (-2, -6), we substitute \(-2\) for x and \(-6\) for y into the equation to see if both sides of the equation remain equal. When performed correctly, point substitution helps in verifying whether a point lies on the curve represented by the equation.
Graph of an Equation
A graph of an equation is a visual representation of all the solutions of that equation. Each solution pair \( x, y \) when plotted on the coordinate axes gives a point on the graph. For linear equations like \( x = 2y - 2 \), the graph will be a straight line. To draw the graph, you can follow these steps:
  • Choose several values for y and calculate the corresponding x values using the equation.
  • Plot the resulting points (x, y) on the coordinate plane.
  • Connect the points to form a straight line.
This visual representation helps in understanding the relationship between x and y. Furthermore, it allows us to easily identify whether a specific point lies on the line by checking if it visually aligns with the plotted points. In our original exercise, the point (-2, -6) does not lie on the graph of the equation \(x = 2y - 2\), which we can confirm both algebraically (through substitution) and visually (on the graph).
Equation Verification
Equation verification is the process of checking whether a given point satisfies the equation of a graph. This involves substituting the point鈥檚 coordinates into the equation and simplifying to see if both sides of the equation are equal. If they are, the point is on the graph; if not, it is not. For our specific problem:
  • We substituted \(x = -2\) and \(y = -6\) into \(x = 2y - 2\).
  • This gave us the equation \(-2 = 2(-6) - 2\).
  • Upon simplifying, we got \(-2 = -12 - 2\) which further simplifies to \(-2 = -14\).
Since \(-2\) is not equal to \(-14\), the point (-2, -6) does not satisfy the equation \(x = 2y - 2\). Thus, the statement that it lies on the graph is false. Following these detailed steps and understanding each process helps in verifying the accuracy of any point relative to a given equation.

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Most popular questions from this chapter

Determine algebraically whether each function is even, odd, or neither. \(F(x)=\sqrt[3]{4 x}\)

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=-x^{2}+3 x-2\)

Determine algebraically whether each function is even, odd, or neither. \(G(x)=\sqrt{x}\)

Explain how the domain of \(g(x)=\sqrt{x}\) compares to the domain of \(g(x-k),\) where \(k \geq 0\).

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the slope and \(y\) -intercept of the line $$3 x-5 y=30$$
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