Problem 12
Question
The graph of each equation is a straight line. Graph the equation after finding the \(x\)-and the \(y\) -intercepts. (since you are given that the graph is a line, you need only plot two points before drawing the line.) $$x=2 y-4$$
Step-by-Step Solution
Verified Answer
The x-intercept is (-4, 0) and the y-intercept is (0, 2).
1Step 1: Rewrite the Equation in Standard Form
Start by rewriting the equation in the form of a standard line equation, which is typically written as \(Ax + By = C\). For the given equation \(x = 2y - 4\), subtract \(2y\) from both sides to get \(x - 2y = -4\).
2Step 2: Find the x-intercept
To find the \(x\)-intercept, set \(y = 0\) in the equation \(x - 2y = -4\). When \(y = 0\), the equation becomes \(x = -4\). Thus, the \(x\)-intercept is \((-4, 0)\).
3Step 3: Find the y-intercept
To find the \(y\)-intercept, set \(x = 0\) in the equation \(x - 2y = -4\). Solving for \(y\) when \(x = 0\) gives \(-2y = -4\); dividing both sides by -2, we get \(y = 2\). Therefore, the \(y\)-intercept is \((0, 2)\).
4Step 4: Plot and Draw the Line
Plot the points for the \(x\)-intercept \((-4, 0)\) and the \(y\)-intercept \((0, 2)\) on the Cartesian plane. Draw a straight line through these points extending in both directions. This line represents the graph of the equation \(x = 2y - 4\).
Key Concepts
x-intercepty-interceptStandard Form of a Linear Equation
x-intercept
When graphing linear equations, finding the x-intercept is a fundamental step. The x-intercept is the point where the graph of an equation crosses the x-axis. At this point, the value of y is always zero. This means to find the x-intercept, you substitute 0 in for y in the equation and solve for x.
For example, in the equation \(x - 2y = -4\), setting \(y = 0\) transforms it to \(x = -4\). Thus, the x-intercept is at the point \((-4, 0)\).
For example, in the equation \(x - 2y = -4\), setting \(y = 0\) transforms it to \(x = -4\). Thus, the x-intercept is at the point \((-4, 0)\).
- To find the x-intercept of any linear equation, set y to zero and solve for x.
- The x-intercept is the point \( (x, 0) \).
y-intercept
The y-intercept is another essential element for graphing linear equations. It is the point where the line crosses the y-axis, which means the x-coordinate at this point is always zero. To find the y-intercept, simply set x to 0 in your equation and solve for y.
Consider our equation \(x - 2y = -4\). By setting \(x = 0\), the equation becomes \(-2y = -4\). Solving for y, you divide both sides by -2, giving \(y = 2\). Therefore, the y-intercept is at \((0, 2)\).
Consider our equation \(x - 2y = -4\). By setting \(x = 0\), the equation becomes \(-2y = -4\). Solving for y, you divide both sides by -2, giving \(y = 2\). Therefore, the y-intercept is at \((0, 2)\).
- To find the y-intercept, set x to zero and solve for y.
- The y-intercept is the point \( (0, y) \).
Standard Form of a Linear Equation
The standard form of a linear equation is a way of writing the equation so it is easily recognizable and workable. A linear equation in standard form looks like \(Ax + By = C\), where A, B, and C are integers, and A should be non-negative. This format is useful because it allows for easy identification of intercepts and graphing.
In our exercise, the original equation was \(x = 2y - 4\). To convert it to standard form, we moved the terms around to rearrange the expression into \(x - 2y = -4\). Here, A is 1, B is -2, and C is -4.
In our exercise, the original equation was \(x = 2y - 4\). To convert it to standard form, we moved the terms around to rearrange the expression into \(x - 2y = -4\). Here, A is 1, B is -2, and C is -4.
- Standard form is useful for quick identification of intercepts.
- The coefficients A, B, and C should ideally be integers to simplify calculations.
Other exercises in this chapter
Problem 11
Evaluate each expression. $$||-8|+|-9 |$$
View solution Problem 12
Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $
View solution Problem 12
Use the distance formula to show that, in each case, the triangle with given vertices is an isosceles triangle. (a) (0,2),(7,4),(2,-5) (b) (-1,-8),(0,-1),(-4,-4
View solution Problem 12
Solve each equation. $$\frac{x}{3}+\frac{2 x}{5}=\frac{-11}{5}$$
View solution