Question

Question #c7d56

Answer 1

`color(red)(y = -5x+15)`

Please refer to the data set chosen, an example, to understand the mathematical process involved in finding the Slope and the intercept of a linear relationship.

Explanation:

We will consider the following data set for our solution:

Table.1

Note: When you graph a linear relationship, the graph is a straight line.

Slope (or) Gradient represents the rate of change of our straight line.

Hence, Slope is represented by a Change in y over a Change in x:

Slope = Change in y`/` Change in x (or)

Slope = `(rise)/(run)` (or)

Slope = `(Delta y)/(Delta x)`

y-intercept is where the graph is going to cross the y-axis.

The Slope-Intercept Equation is given by the formula

`color(blue)(y = mx + b)` `color(red)(..Equation.1)`

where `color(blue)(m)` is our Slope and `color(blue)(b)` is our y-intercept

`color(green)(Step.1)`

in this step, we will now investigate for the Change in y in our table of values available in Table 1

What happens when we move from the first value of `y` to the second value of `y`?

Table.2

We can see that the differences are `color(blue)(-5)` among the y-values.

Hence `(Delta y) = -5` `color(red)(..Equation.2)`

`color(green)(Step.2)`

In this step, we will find out whether there is a Constant Rate of Change for our x-values

Table.3

Observe that we do have a constant rate of change for our x-values.

Hence `(Delta x) = 1` `color(red)(..Equation.3)`

`color(green)(Step.3)`

In this step, we are ready to find our Slope

Using our equations `color(red)(Equation.2 and Equation.3)` we can write

Slope = `(Delta y)/(Delta x)`

Slope `= -5/1 = -5`

Hence, our Slope(m) = -5... Result.1

`color(green)(Step.4)`

Our y-intercept is when our graph (in our example, it is a straight line) is crossing the y-axis

We know that when our graph crosses the y-axis our `x` value will be zero(0)

From our Table.1 , we understand that `x = 0` when the corresponding `y.value = 15`

Hence, our y-intercept = 15 ... Result.2

It means that this is the point on our y-axis where our graph will cross through.

`color(green)(Step.5)`

In this step, we are ready to write the Equation of our linear relationship, in the slope-Intercept Form

Using `color(red)(..Equation.1)` and our intermediate results, Result.1 and Result.2 we obtain

`color(blue)(y = mx + b)`

`color(red)(y = -5x+15)` our final answer.