Problem 58
Question
Graph each equation in a rectangular coordinate system. $$3 x+12=0$$
Step-by-Step Solution
Verified Answer
The Graph of the equation \(3x + 12 = 0\) in a rectangular coordinate system is a straight line parallel to the y-axis passing through the point \((-4,0)\)
1Step 1: Solve the Equation
First, arrange the equation by sliding the constant term to the other side to determine the value of \(x\). So, \(3x + 12 = 0\) becomes \(3x = -12\). To find the value of \(x\), divide both sides by the coefficient 3. Doing this results in \(x = -4\).
2Step 2: Plotting on the Graph
Now, plot this point on the \((x, y)\) plane. As it's a linear equation, the straight line passing through the point \((-4,0)\) and parallel to the y-axis represents this equation. This is because the equation is independent of \(y\), hence, for every value of \(y\), \(x\) will always be -4.
Key Concepts
Rectangular coordinate systemLinear equationsVertical line graphEquation solving steps
Rectangular coordinate system
The rectangular coordinate system, also known as the Cartesian coordinate system, is a two-dimensional plane used for graphing equations. It consists of two perpendicular lines, or axes: the horizontal axis known as the x-axis, and the vertical axis known as the y-axis. These axes intersect at a point called the origin, denoted as (0,0). This system helps us place points on a plane using pairs of numbers called coordinates.
Each point is represented by \(x, y\), where \(x\) is the horizontal distance from the origin and \(y\) is the vertical distance. This system allows us to visually represent equations and understand their behavior easily by plotting the points or lines that satisfy them.
Each point is represented by \(x, y\), where \(x\) is the horizontal distance from the origin and \(y\) is the vertical distance. This system allows us to visually represent equations and understand their behavior easily by plotting the points or lines that satisfy them.
Linear equations
Linear equations are equations that form straight lines when graphed on a coordinate system. They have the general form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. These equations show a direct relationship between the variables \(x\) and \(y\).
In many cases, you solve for one variable in terms of the other to draw the graph. For example, solving for \(x\) will tell you how \(x\) changes with \(y\). This is central in analyzing the slope and intercepts of lines on a graph. Linear equations help us quickly identify relationships in equations just by visual inspection, making them advantageous for problem-solving.
In many cases, you solve for one variable in terms of the other to draw the graph. For example, solving for \(x\) will tell you how \(x\) changes with \(y\). This is central in analyzing the slope and intercepts of lines on a graph. Linear equations help us quickly identify relationships in equations just by visual inspection, making them advantageous for problem-solving.
Vertical line graph
A vertical line graph results from an equation in which the x-value is constant, as seen in the equation \(3x + 12 = 0\). Solving this gives \(x = -4\), meaning for any \(y\)-value, \(x\) remains \(-4\).
This type of line runs parallel to the y-axis and indicates no relationship between \(x\) and \(y\); hence, there's no slope. You plot it by drawing a vertical line at the constant x-value. It's useful in identifying constraints or fixed conditions within real-world scenarios.
This type of line runs parallel to the y-axis and indicates no relationship between \(x\) and \(y\); hence, there's no slope. You plot it by drawing a vertical line at the constant x-value. It's useful in identifying constraints or fixed conditions within real-world scenarios.
Equation solving steps
Solving equations involves finding the value of the variable that makes the equation true. Let's look at \(3x + 12 = 0\).
- First, rearrange the equation by moving constants to one side, which transforms \(3x + 12 = 0\) into \(3x = -12\).
- Next, isolate the variable by dividing all terms by the constant coefficient, giving \(x = -4\).
Other exercises in this chapter
Problem 58
Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=x^{2}+1, g(x)=x^{2}-3$$
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Begin by graphing the standard quadratic function, \(f(x)=x^{2} .\) Then use transformations of this graph to graph the given function. $$h(x)=-(x-1)^{2}$$
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complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}-2 x+y^{2}-15=0 $$
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Let $$\begin{aligned}&f(x)=2 x-5\\\&g(x)=4 x-1\\\&h(x)=x^{2}+x+2\end{aligned}$$ Evaluate the indicated function without finding an equation for the function. $$
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