I have this question as part of my Pre-Calc summer work. I never learned about f(x) equations without the variable x in them... it just doesn't make sense to me based on what I know. Please help! I need to turn this thing in today if possible..
How do you put f(-2)=1 and f(-1)=3 in slope-intercept form?
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:
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.
Solving the system of equations for m and b we get
m=2 and b=5
"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 :)