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Chapter 1: Problem 12

Find an equation of the given line. \(\left(\frac{1}{2}, 1\right)\) and \((1,4)\) on line

Short Answer

Expert verified
The equation of the line is \(y = 6x - 2\).

Step by step solution

01

- Find the slope

The slope (m) of a line passing through two points \(x_1, y_1\) and \(x_2, y_2\) can be found using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] For the points \(\left(\frac{1}{2}, 1\right)\) and \(\left(1, 4\right)\), we have: \[ m = \frac{4 - 1}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = \frac{3}{0.5} = 6 \]
02

- Use point-slope form

The point-slope form of a line's equation is \[ y - y_1 = m(x - x_1) \] Using the point \(\left(\frac{1}{2}, 1\right)\) and the slope \(m = 6\), we get: \[ y - 1 = 6(x - \frac{1}{2}) \]
03

- Simplify the equation

Distribute the slope and simplify: \[ y - 1 = 6(x - \frac{1}{2}) \] \[ y - 1 = 6x - 3 \] Adding 1 to both sides, we get: \[ y = 6x - 2 \]

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Key Concepts

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

Slope of a Line
To determine the equation of a line, the first step is finding its slope. The slope represents how steep the line is and is denoted by the symbol \(m\). You can calculate the slope if you have two points on the line.
Use this formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of the two points.
In our example, the points are \(\frac{1}{2}, 1\) and \(1, 4\). Plugging these values into the formula gives:
\[ m = \frac{4 - 1}{1 - \frac{1}{2}} = \frac{3}{0.5} = 6 \]
Therefore, the slope of our line is 6.
The slope helps us understand how much the y-value changes for each unit increase in the x-value.
Point-Slope Form
Once we know the slope, we can use the point-slope form to write the equation of the line. This form is especially useful because it incorporates the slope and one point from the line.
The point-slope form equation is:
\[ y - y_1 = m(x - x_1) \]
Here, \(x_1, y_1\) are the coordinates of the point on the line, and \(m\) is the slope.
In this case, we will use the point \(\frac{1}{2}, 1\) and our calculated slope of 6:
\[ y - 1 = 6(x - \frac{1}{2}) \]
This equation tells us how y, the dependent variable, changes with x.
Linear Equations
Linear equations represent lines on a graph. These equations can be simplified into the slope-intercept form:
\[ y = mx + b \]
Where \(m\) is the slope, and \(b\) is the y-intercept (the point where the line crosses the y-axis).
To convert our point-slope form to the slope-intercept form, we distribute and simplify:
\[ y - 1 = 6(x - \frac{1}{2}) \] \[ y - 1 = 6x - 3 \] Adding 1 to both sides gives:
\[ y = 6x - 2 \]
This final equation \(y = 6x - 2\) is in slope-intercept form and shows that our line has a slope of 6 and crosses the y-axis at -2.

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Most popular questions from this chapter

(a) Let \(A(x)\) denote the number (in hundreds) of computers sold when \(x\) thousand dollars is spent on advertising. Represent the following statement by equations involving \(A\) or \(A^{\prime}:\) When $$\$ 8000$$ was spent on advertising, the number of computers sold was 1200 and it was rising at the rate of 50 computers for each $$\$ 1000$$ spent on advertising. (b) Estimate the number of computers that will be sold if $$\$ 9000$$ is spent on advertising.

Compute the limits that exist, given that $$ \lim _{x \rightarrow 0} f(x)=-\frac{1}{2} \text { and } \lim _{x \rightarrow 0} g(x)=\frac{1}{2} \text { . } $$ (a) \(\lim _{x \rightarrow 0}(f(x)+g(x))\) (b) \(\lim _{x \rightarrow 0}(f(x)-2 g(x))\) (c) \(\lim _{x \rightarrow 0} f(x) \cdot g(x)\) (d) \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}\)

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. \(f^{\prime}(2)\), where \(f(x)=\frac{x}{1+x}\)

Use paper and pencil to find the equation of the tangent line to the graph of the function at the designated point. Then, graph both the function and the line to confirm it is indeed the sought-after tangent line. \(f(x)=x^{3},(1,1)\)

Determine whether each of the following functions is continuous and/or differentiable at \(x=1\). \(f(x)=\left\\{\begin{array}{ll}x & \text { for } x \neq 1 \\ 2 & \text { for } x=1\end{array}\right.\)
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