Page 3.3: More on Linear Functions
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)At this point in time, we should hopefully feel confident in working with functions. The goal now is to discuss linear functions in more depth. Although we did touch upon linear functions in Section 2.5, there is a lot more to note!
Recall that a linear function is a function \(f: \mathbb{R} \rightarrow \mathbb{R} \) defined by \( f(x) = mx + b \) or \( y = mx + b \). We refer to the form \( y = mx+b \) is as the slope-intercept form (can you explain why?).
As a quick review exercise, take a few moments to graph a linear functions that meets the following conditions.
1) When the slope is positive and \(y\)-intercept is positive.
2) When the slope is positive and the \(y\)-intercept is negative.
3) When the slope is positive and the \(y\)-intercept is zero.
4) When the slope is negative and the \(y\)-intercept is positive.
5) When the slope is negative and the \(y\)-intercept is negative.
6) When the slope is negative and the \(y\)-intercept is zero.
7) When the slope is zero and the \(y\)-intercept is positive.
8) When the slope is zero and the \(y\)-intercept is negative.
9) When the slope is zero and the \(y\)-intercept is zero.
Finding the Equation of a Line Given Two Points
Consider two points on the \(xy\)-plane as shown below:
How many linear lines can you draw which connects these two points?
- Answer
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We see that there is only one such linear line that connects these two points.
The question we will now consider is given two points, how can we find this single, unique, linear line, \(y = mx + b\), that connects the two points? We see that to find the line, we must determine the value for \(m\) and the value for \(b\).
To find the linear line connecting two points, \( (x_1, y_1), (x_2, y_2) \) we first:
1) Find the slope, \(m\) by computing \( m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{rise}{run} = \dfrac{ \Delta y}{ \Delta x} \).
2) With the value of \(m\) now known, we plug in one of the points \( (x_1, y_1) \) or \( (x_2, y_2) \) into \(y = mx + b\) to find the value of \(b\).
Find the equation of the linear line that passes through the points \( (3,1) \) and \( (7,4) \).
- Answer
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1) We first compute \(m\):
\begin{align*} m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4-1}{7-3} = \frac{3}{4} \end{align*}
2) We now know that \( m = \frac{3}{4} \) and we shall select any one of the given points, it does not matter which one we select. Suppose we selected \( (3,1) \). Plugging in \( m \) and the point \( (3,1) \) into the equation \( y = mx + b \) yields:
\begin{align*} y &= mx + b \\ \\ 1 &= \dfrac{3}{4} (3) + b \\ \\ 1 &= \frac{9}{4} + b \\ \\ - \frac{5}{4} &= b \end{align*}
Hence \( y = \frac{3}{4} x - \frac{5}{4} \).
To check our answer, we plug in the \(x\)-coordinate of both points to make sure we obtain the appropriate value of \(y\).
Find the equation of the line which passes through the points:
1) \( (2,1) \) and \( (-1,-5) \).
2) \( (-5,7) \) and \( (5,9) \).
3) \( (0,3) \) and \( (4,15) \).
4) \( (2,7) \) and \( (4,7) \).
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1) \( y = 2x-3 \)
2) \( y = \frac{1}{5} x + 8 \)
3) \( y = 3x +3 \)
4) \( y = 7 \)
An important feature regarding linear lines is that given a linear line, regardless of which two (distinct) points are chosen, the slope will always be the same.
Finding the Equation of a Line Given a Point and a Slope
In the above discussion, we said that we can determine the equation of the linear line given two points on the line. However, we can also determine the equation of a linear line given a point and a slope. Essentially, this is just step 2 we are performing in the above recipe.
Find the equation of the line passing through the point \( (-1,4) \) and whose slope is \(10\).
Solution 1
Since the equation of a line is given by \(y = mx + b \) and we know that \(m = 10 \), then we have the following:
\begin{align*} y &= mx + b \\ \rightarrow 4 &= 10(-1) + b \\ 4 &= -10 + b \\ 14 &= b \end{align*}
Hence \( y = 10x + 14 \).
Alternatively, we can use what is called the point-slope form of a line which we will often seen in calculus.
Definition: The point-slope form of a linear line is given by the following equation: \( y -y_1 = m(x-x_1) \).
Although many students have seen the above equation, they are often unsure as to where it comes from. Imagine that we only have one point \( (x_1, y_1) \) and \(m\). Recall that the slope formula is given by
\[ m = \dfrac{y_2 - y_1}{x_2 - x_1} \nonumber\ \]
Since we only have one point, allow us to drop the second index that appears in the above formula. Doing so gives us:
\[ m = \dfrac{y - y_1}{x - x_1} \nonumber\ \]
Multiplying both sides by \( x - x_1 \) yields \[ y -y_1 = m(x-x_1) \nonumber\ \] which is the point-slope formula.
Allow us to use the point-slope formula to answer the same question above:
Find the equation of the line passing through the point \( (-1,4) \) and whose slope is \(10\).
- Answer 2
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From the point-slope formula, we have
\begin{align*} y -y_1 &= m(x-x_1) \\ \\ y - 4 &= 10 \bigg(x - (-1) \bigg) \\ \\ y - 4 &= 10 (x+1) \\ \\ y - 4 &= 10x + 10 \\ \\ y - 4 &= 10x + 10 \\ \\ y &= 10x + 14 \end{align*}
Hence \( y = 10x + 14 \).
Finally, we end this section by discussing a particular type of line which is not a function. Recall that the line determined by \( y = c \) is indeed a function since this is a horizontal line that passes through \(c\) on the \(y\)-axis. However, it's counterpart, which is a vertical line, is not a function.
The graph of the equation \(x = a \) is a vertical line which passes through \(a\) on the \(x\)-axis.
Select any two points on the line \( x = 2 \) and compute the slope. What do you notice?
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Suppose we selected the points \( (2,1) \) and \( (2,5) \). We see that if we were to use the slope formula, we would obtain:
\[ m = \frac{5-1}{2-2} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4}{0} = \text{not defined} \nonumber\ \]
The same is true for any two points on a vertical line and so we note that the slope of any vertical line is not defined. (Note that some people say that the slope is infinite and the reason for this statement will be explained on Day 7).