Determine the equation of the following linear functions.
f goes by A\left(-2,-1\right) and B\left(2{,}4\right).
First, we need to find the slope of f. If f goes by A\left(x_A,y_A\right) and B\left(x_B,y_B\right), then the slope of f is:
m= \dfrac{y_B-y_A}{x_B-x_A}
Here, f goes by A\left(-2,-1\right) and B\left(2{,}4\right) . We have:
m= \dfrac{4-\left(-1\right)}{2-\left(-2\right)}= \dfrac{5}{4}
Furthermore, when f goes by \left(a,b\right), one writes:
f\left(x\right)-b = m\left(x-a\right)
Since f goes by B\left(2{,}4\right), we may write:
f\left(x\right)-4 = \dfrac{5}{4}\left(x-2\right)
f\left(x\right) = \dfrac{5}{4}x-\dfrac{10}{4}+4
f\left(x\right) = \dfrac{5}{4}x + \dfrac{3}{2}
f passes through \left(-1{,}1\right) and (1,0).
The slope of f is:
m= \dfrac{y_B-y_A}{x_B-x_A}
It can be computed as follows:
m= \dfrac{0-1}{1-\left(-1\right)}= \dfrac{-1}{2}
Furthermore, when f goes by \left(a,b\right), one writes:
f\left(x\right)-b = m\left(x-a\right)
Since f goes through B\left(1{,}0\right), we may write:
f\left(x\right)-0 = -\dfrac{1}{2}\left(x-1\right) \\f\left(x\right)=-\dfrac{1}{2}x+\dfrac{1}{2}
f\left(x\right)=-\dfrac{1}{2}x+\dfrac{1}{2}
f passes through \left(3{,}2\right) and (3,-4)
Since all the x -coordinates are the same, we deduce that f is parallel to the y -axis. We can immediately say that the equation of the line is x=3.
x=3
f goes by \left(3{,}0\right) and \left(0,-2\right)
The slope of f is:
m= \dfrac{y_B-y_A}{x_B-x_A}
It equals:
\dfrac{-2-0}{0-3} = \dfrac{2}{3}
Furthermore, when f goes by \left(a,b\right), one writes:
f\left(x\right)-b = m\left(x-a\right)
We can write:
f\left(x\right) - 0 = \dfrac{2}{3}\left(x-3\right)\\ \Rightarrow f\left(x\right) = \dfrac{2}{3}x-2
f\left(x\right) = \dfrac{2}{3}x-2
f passes through \left(1,-1\right) and \left(-2,-1\right)
Since the second coordinates are equal, we conclude that the line is parallel to the x -axis. Therefore, the equation of the line is f\left(x\right) = -1.
f(x) = -1
f goes through \left(-1{,}1\right) and \left(1,-1\right)
The slope of f is:
m= \dfrac{y_B-y_A}{x_B-x_A}
It equals:
\dfrac{-1-1}{1-\left(-1\right)}= \dfrac{-2}{2}=-1
Furthermore, when f goes by \left(a,b\right), one writes:
f\left(x\right)-b = m\left(x-a\right)
So, we may write:
f\left(x\right)-1 = -1\left(x-\left(-1\right)\right) \Rightarrow f\left(x\right) 1=-x-1 \Rightarrow f\left(x\right)=-x
f\left(x\right)=-x
f passes through \left(1{,}2\right) and \left(2{,}3\right)
The slope of f is:
m= \dfrac{y_B-y_A}{x_B-x_A}
So it equals:
\dfrac{2-1}{3-2}=1
Furthermore, when f goes by \left(a,b\right), then one writes:
f\left(x\right)-b = m\left(x-a\right)
Therefore, we have:
f\left(x\right) - 2= 1\left(x-1\right) \Rightarrow f\left(x\right) = x+1
f\left(x\right) = x+1