Linear function through two points

A way to do this, which was not seen in high school, is to sum up a function $g(x)$ going through $(2, 0)$ and $(3, -2)$ with another function $h(x)$ going through $(2, 4)$ and $(3, 0)$ .

In other words:

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With a bit of math and logic, you can notice that if a function goes through points $(x_0, 0)$ and $(x_1, y_1)$ , then it will be expressed as:

Knowing that $m=\frac{y_1}{x_1-x_0}$ is the slope and that $x_0$ outputs as $0$ .

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Now that we identified the two functions, you get:

• $g(x)=\frac{-2}{3-2}\cdot (x-2))$

• $h(x)=\frac{4}{2-3}\cdot (x-3))$

$f(x)=g(x)+h(x)$

$= ( \frac{-2}{3-2}\cdot (x-2)) + ( \frac{4}{2-3}\cdot (x-3))$

$= -2(x-2)-4(x-3)$

$= -2x+4-4x+12$

We can determine the gradient of our line by using the formula $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$ : $m = \cfrac{-2-4}{3-2} = \cfrac{-6}{1} = \boxed{-6}$ Now, our equation for our line looks like $y=-6x+c$ but we need to find $c$ . For this, we can use one of our coordinates (I used the first one) to determine the value of our unknown constant: $\begin{aligned} y=-6x+c \xRightarrow{\text{sub in }x=2\text{ and }y=4} 4 &= -6(2) + c \\ \implies 4 &= -12 + c \\ \implies c &= 4 + 12 = \boxed{16} \end{aligned}$ Now we can form our equation and it is $\green{\boxed{y=-6x + 16}}$

The equation for a straight line is given by $f(x) = y = mx + c$ , where $m$ is the gradient and $c$ is the $y$ -intercept of the straight line. Since the two points $(2,4)$ and $(3,-2)$ satisfy the equation, we have:

$\begin{cases} \begin{aligned} 4 = 2m + c & ...(1) \\ -2 = 3m + c & ...(2) \end{aligned} \end{cases} \implies (2)-(1): \ -6 = m$ . Putting $m=-6$ in $(1)$ , $\implies 4 = -12 + c \implies c = 16$ .

Therefore, $f(x) = \boxed{-6x+16}$ .