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write the equaion of a line in slope-intercept form if the slope is 1/5 and the y-intercept is -9 explain please

Sagot :

Answer:

[tex]y = \frac{1}{5} x - 9[/tex]

Step-by-step explanation:

The slope-intercept form is y = mx + b, where:

m = slope

b = y-intercept.

The slope (m) tells you the steepness of the line. It is the average rate of change which measures how the y-value changes for each one-unit change in the x-value. Hence, slope [tex]= \frac{rise}{run}[/tex].  So the given slope of 1/5 means that for every 1 unit change in the y-value, the x-value changes by 5 units (you go up 1 unit, and "run" 5 units to the right).

Next, the y-intercept is the point on the graph where it crosses the y-axis, and has the coordinates, (0, b). It is also the value of y when x = 0. Since you're given the y-intercept of -9, then that means that it is the y-coordinate of (0, b). So, it becomes (0, -9).

Now that we have our slope (m ) = 1/5, and the y-intercept (b ) = -9, we can write the equation of the line as:

[tex]y = \frac{1}{5} x - 9[/tex]

(I'm also including a screenshot of the line where it shows the y-intercept of (0, -9) on the graph).  

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View image djtwinx017