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How do you put f(-2)=1 and f(-1)=3 in slope-intercept form?

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4 Answers

Let's just think of f(x) as another way to refer to y

f(x) = y

f(-2) = 1, is telling you that when x = -2, y = 1

f(-1) = 3, is telling you that when x = -1, y = 3

So they're giving you 2 points and asking you to find the equation of the line.

(-2, 1) (-1, 3)

To go from here, you can find the slope (change in y over change in x) and then write out the equation using point-slope form. Lastly, rearrange into slope intercept form and you're all set.

Check out this site of an example and step by step directions:

http://www.mathsisfun.com/algebra/line-equation-2points.html

The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. If we know the positions of two points on a line, then we can know both the line's slope and y-intercept. Let me explain.

So you've got two points on the line, (-2, 1) and (-1, 3). We can calculate the slope by rise over run, so m = (3 - 1)/((-1) - (-2)) or 2. To get b, we can plug the x- and y-values of either point; say we do it for the first point, then what we get is 3 = 2(-1) + b; from this, we can get that b = 5.

Hence the line's equation is y = 2x + 5.

I like the previous answer.  Alternatively...

the slope intercept form for a straight line is y=mx+b (m is slope and b the y intercept)

This can also be written as f(x)=mx+b by convention.  f(-2)=-2m+b.  

f(-2)=1=>-2m+b=1

f(-1)=3=>-m+b=3

Solving the system of equations for m and b we get 

m=2 and b=5

Hi Kylie!

"f(-2) means we're using x= -2 ... thus f(-2)=1 lands at the point (-2,1), f(-1)=3 lands at (-1,3) ... looks like the "rise over run" is 2/1 or slope m= 2 ... find b: 1= 2(-2) +b ... b=5 ... y= 2x +5 works ... Regards :)