4.1 Linear functions  (Page 5/27)

If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y -axis when the output value is 7. Therefore, $b=7.$ We now have the initial value $b$ and the slope $m$ so we can substitute $m$ and $b$ into the slope-intercept form of a line.

So the function is $f(x)=-\frac{3}{4}x+7,$ and the linear equation would be $y=-\frac{3}{4}x+7.$

Writing an equation for a linear function

Write an equation for a linear function given a graph of $f$ shown in [link] .

Identify two points on the line, such as $(0,2)$ and $(\mathrm{-2},\mathrm{-4}).$ Use the points to calculate the slope.

$\begin{array}{ccc}\hfill m& =& \frac{y_2-y_1}{x_2-x_1}\hfill \\ & =& \frac{\mathrm{-4}-2}{\mathrm{-2}-0}\hfill \\ & =& \frac{\mathrm{-6}}{\mathrm{-2}}\hfill \\ & =& 3\hfill \end{array}$

Substitute the slope and the coordinates of one of the points into the point-slope form.

$\begin{array}{ccc}\hfill y-y_1& =& m(x-x_1)\hfill \\ \hfill y-(\mathrm{-4})& =& 3(x-(\mathrm{-2}))\hfill \\ \hfill y+4& =& 3(x+2)\hfill \end{array}$

We can use algebra to rewrite the equation in the slope-intercept form.

$\begin{array}{ccc}\hfill y+4& =& 3(x+2)\hfill \\ \hfill y+4& =& 3x+6\hfill \\ \hfill y& =& 3x+2\hfill \end{array}$

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Modeling real-world problems with linear functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.