# How do you find the slope given (3,3) and (4,0)?

May 13, 2018

The slope in the function f(x)=ax+b, where f(x) goes through (3, 3) and (4, 0) is $a = - 3$

#### Explanation:

We need to start by making a graph. It makes it easier to see what we have:

(By the way, it is important that you don't write $\left(3 , 3\right)$ and $\left(4 , 0\right)$ as these can be read as the decimal numbers $3.3$ and $4.0$, which at least some non English speaking languages use. To ensure no misunderstanding, therefore, please add a space after the comma: $\left(3 , 3\right)$, $\left(4 , 0\right)$)

As we can see, this is a linear function on the form
$y = a x + b$, where the slope is the constant $a$.
(Alternatively we can write it as $f \left(x\right) = a x + b$, but it is slightly easier to handle the function if we use $y$.)

From the graph we can read directly that $a = - 3$, since if x increases with 1 from 3 to 4, y falls from 3 to 0 in value.

We can show this the following way: The two value pairs $x = 3 , f \left(x\right) = 3$ and $x = 4 , y = 0$ both fulfill the function
$y = a x + b$ which we are interested in.

If we plug these two value pairs into the function, we get the following pair of equations:
(1) $3 = 3 a + b \implies 3 a + b = 3$
(2) $0 = 4 a + b \implies 4 a + b = 0$

As we have $+ b$ in both, we subtract (1) from (2) to get rid of it:

$\left(4 a + b\right) - \left(3 a + b\right) = 0 - 3$
$\implies a = - 3$
Our slope is $- 3$

$4 a + b = 0$ it follows that $b = - 4 a = \left(- 4\right) \cdot \left(- 3\right) = 12$