Linear functions describe functions that preserve both additivity and scalar multiplication.
To break this jargon down:
To think about this much more easily, Linear Functions are functions where the input variable has a degree (that is, exponent value) of 1 (like f(x)=2x + 2) or 0 (like f(x)=2).
Linear functions present themselves like the following chart. Notice that the graph is just a line (hence, LINEar)
Constant functions are a special type of linear function where, for all output, the output is the same value. These are of the form f(x) = c, where c simply represents some numerical value.
Identity: A special type of linear function where the output is the same as the input. It is of the form, f(x) = x.
Linear functions have important characteristics beyond the preservation of additivity and scalar multiplication.
Linear functions are typically of the form:
y = mx + b
Where y is the output variable (could be f(x), g(x), z, etc.), x is the input variable, and:
For this reason, the form, y=mx+b is known as slope-intercept form since it describes both the slope and the intercept of a single function.
Another form that may be utilized is the point-slope form, which enables defining a function utilizing two ordered pairs.
y - y_1 = m(x-x_1)
The first pair is going to be (x, y), as you are defining the behavior between x and y. The second pair will be ANY point on the line, represented by (x_1, y_1).
Point-slope form may always be simplified into slope-intercept form, and slope-intercept form may always be expanded into point-slope form (so long as the point chosen resides on the line).
Example:
y - 3 = 2(x-4)
y - 3 = 2x - 2(4)
y - 3 = 2x - 8
\boxed{y = 2x - 5}
Point-slope form is converted to slope-intercept by solving for y!
Slope-intercept form is more open-ended, since you may chose any point. You simply extract the slope and choose a point on the line. For instance: y = f(x) = 2x - 5.
Say we choose to utilize x_1=12.
First, solve for f(12): y = f(12) = 2(12) - 5 = 24 - 5 = \boxed{19}
This is your (x_1, y_1) ordered pair. Now, we know that the original slope is 2 (since, in slope intercept form, we have y = \underline{2}x - 5). Plug this into the point-slope form template:
y - y_1 = m(x - x_1)
y - 19 = 2(x - 12)
To confirm there are no more mistakes, you can convert it back to slope intercept form, to ensure that the original form is the same as this new form!
y - 19 = 2(x - 12)
y - 19 = 2x - 24
\boxed{y = 2x - 5} \; \checkmark
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