A line passes through the points,
$\text{\hspace{0.17em}}(\mathrm{-2},\text{\u221215})\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}(2,\mathrm{-3}).\text{\hspace{0.17em}}$ Find the equation of a perpendicular line that passes through the point,
$\text{\hspace{0.17em}}(6,4).$
Linear functions can be represented in words, function notation, tabular form, and graphical form. See
[link] .
An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See
[link] .
Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between
y -values by the difference in corresponding
x -values of any two points on the line. See
[link] and
[link] .
An equation for a linear function can be written from a graph. See
[link] .
The equation for a linear function can be written if the slope
$\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ and initial value
$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ are known. See
[link] and
[link] .
A linear function can be used to solve real-world problems given information in different forms. See
[link],[link], and
[link] .
Linear functions can be graphed by plotting points or by using the
y -intercept and slope. See
[link] and
[link] .
Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See
[link] .
The equation for a linear function can be written by interpreting the graph. See
[link] .
The
x -intercept is the point at which the graph of a linear function crosses the
x -axis. See
[link] .
Horizontal lines are written in the form,
$\text{\hspace{0.17em}}f(x)=b.\text{\hspace{0.17em}}$ See
[link] .
Vertical lines are written in the form,
$\text{\hspace{0.17em}}x=b.\text{\hspace{0.17em}}$ See
[link] .
Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See
[link] .
A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the
x - and
y -values of the given point into the equation,
$\text{\hspace{0.17em}}f(x)=mx+b,\text{\hspace{0.17em}}$ and using the
$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ that results. Similarly, the point-slope form of an equation can also be used. See
[link].
A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See
[link] and
[link] .
Section exercises
Verbal
Terry is skiing down a steep hill. Terry's elevation,
$\text{\hspace{0.17em}}E(t),$ in feet after
$\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by
$\text{\hspace{0.17em}}E(t)=3000-70t.\text{\hspace{0.17em}}$ Write a complete sentence describing Terry’s starting elevation and how it is changing over time.
Terry starts at an elevation of 3000 feet and descends 70 feet per second.
Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?
A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after
t hours.
not much
For functions, there are two conditions for a function to be the inverse function:
1--- g(f(x)) = x for all x in the domain of f
2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎