Problem 33
Question
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through \((-3,-1)\) and \((4,-1)\)
Step-by-Step Solution
Verified Answer
The slope of the line between points \((-3,-1)\) and \((4,-1)\) equals 0 suggesting a horizontal line. The point-slope form of the line equation is \(y + 1 = 0(x + 3)\) simplifying to \(y = -1\). The slope-intercept form is also \(y = -1\), 'c' equals -1.
1Step 1: Calculating the Slope of the Line
Calculate the slope of the line using the given points \((-3,-1)\) and \((4,-1)\) using the formula \(m=(y_2-y_1)/(x_2-x_1)\).
2Step 2: Writing the Point-Slope Form Equation
Now use the slope and one of the points, say \((-3,-1)\) to write the point-slope form of the equation as \(y - y_1 = m(x - x_1)\).
3Step 3: Converting to the Slope-Intercept Form Equation
Transform the point-slope form into the slope-intercept form using algebraic manipulation, achieving \(y = mx + c\). You can consider that for a horizontal line \(c = y_1\).
4Step 4: Finding the y-intercept c
Use the slope-intercept form and one of the points, again let’s use \((-3,-1)\), to solve for c.
Key Concepts
Point-Slope FormSlope-Intercept FormSlope CalculationHorizontal Line Equation
Point-Slope Form
The point-slope form is a popular way to write the equation of a line when you know a specific point on the line and the slope. It's expressed as \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is a point on the line, and \( m \) is the slope. This form is very handy because it clearly shows how the line passes through the given point.
- Choose any point, like \((-3, -1)\) from the original problem.
- Once you calculate the slope, insert it into the equation along with the chosen point.
Slope-Intercept Form
The slope-intercept form of a line is one of the most common ways to represent a linear equation. It's easy to understand as it highlights both the slope and the y-intercept directly in its form: \[ y = mx + c \] where \( m \) stands for slope and \( c \) stands for the y-intercept. This form allows for quick graphing since once \( m \) and \( c \) are known, you can see where the line crosses the y-axis and its steepness.
- After finding the slope, manipulate the point-slope form to this slope-intercept form.
- Plug in any known point to solve for \( c \), making it possible to completely frame the equation.
Slope Calculation
Calculating the slope is an essential first step when writing the equation of a line. Slope, denoted by \( m \), measures how steep the line is. The formula used to calculate slope given two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
- This formula calculates the change in \( y \) (rise) over the change in \( x \) (run).
- For the given points in the problem, since both points \((-3, -1)\) and \((4, -1)\) have the same \( y \)-value, you will find that \( m = 0 \).
Horizontal Line Equation
A horizontal line is special because its slope is always zero. This directly influences its equation. For any horizontal line, the equation takes a simple form: \[ y = c \] where \( c \) is the constant y-value across the entire line.
- This is because no matter what \( x \) value you choose, \( y \) will always remain the same.
- From the given points, since \( y \) is \(-1\) for both \(-3\) and \(4\), the line is \( y = -1 \).
Other exercises in this chapter
Problem 32
Write the standard form of the equation of the circle with the given center and radius. $$ \text { Center }(0,0), r=8 $$
View solution Problem 33
Begin by graphing the absolute value function, \(f(x)=|x| .\) Then use transformations of this graph to graph the given function. $$ h(x)=2|x+4| $$
View solution Problem 33
Find: a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=\sqrt{x}, g(x)=x+3$$
View solution Problem 33
In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=4 x$$
View solution