# what is the linear function y=mx+c if points (-1,6) and (2,15) are on the graph of function?

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First, calculate the slope using the formula

m = (y2 - y1) / (x2 / x1)

m = (15 - 6) / (2 - -1)

m = 9 / 3

m = 3

Substitute 3 in for m in the equation y = mx + b. Select one of the points and substitute these values in for x and y. Then solve for b.

y = mx + b

6 = 3 * -1 + b

6 = -3 + b

9 = b

Therefore, the equation is...

**y = 3x + 9**

Written as a function...

**f(x) = 3x + 9**

To determine the linear function, we'll apply the following formula:

y - y1 = m(x - x1), where m is the slope of the line and (x1,y1) are the coordinates of one of two given points.

m = (y2-y1)/(x2-x1)

Let x1 = -1 , y1 = 6 , x2 = 2 , y2 = 15

m = (15-6)/(2+1)

m = 9/3

m = 3

We'll plug in the values of m and x1 and y1 into the formula that gives the equation of the line.

y - 6 = 3(x + 1)

We'll put the equation of the line into the requested slope intercept form. For this reason, we'll have to isolate y to the left side.

y - 6 = 3x + 3

y = 3x + 3 + 6

y = 3x + 9

**The requested equation of the line is y = 3x + 9.**