How do I find the function of a line knowing 3 points?

Question

I've got 3 points on a line, and I would like to know the function. If possible, I'd like to do it algebraically, but if that's too hard I can live with a solution using technology. My TI-84+ broke so I can't do it myself, will you please help me?

The three points are (2sin(-pi/8),0.25), (2,0.5) and the origin. The domain is between 0 and 2, and it looks like it decelerates.

Answer 1

To find the function of a LINE you need only two points.

First you find the slope using two points, say (2, 0.5) and (0,0)

m = (0.5-0)/(2-0) = 1/4

Then you solve

(y-y1) = m(x-x1) to find the value for b and get the form y = mx + b, where (x1,y1) is one of the points.

Lets use (x1,x2) = (0,0) --- it's the easiest

(y-0) = 1/4 (x-0)

y = (1/4) x

Now the problem you have is that point (2sin(-π/8), 0.25) doesn't fit on that line.

So perhaps you are looking for the function of a curve, i.e.

y = ax^2 + bx + c ??

Start by substituting x and y by (0,0) to find c

0 = a(0) + b(0) + c

c= 0

So now we have

y = ax^2 + bx

Substituting in the other two points, we get:

Point (2, 0.5):

0.5 = a(2)^2 + b(2)

4a + 2b = 0.5

Point (2sin(-π/8),0.25):

0.25 = a(2sin(-π/8))^2 + b(2sin(-π/8))

4(sin(-π/8))^2 a + 2sin(-π/8) b = 0.25

Now you have two equations with two unknowns a, b

NOW, for the problem with your question:

You say that the domain (x-values) is between 0 and 2, and yet

(2sin(-pi/8),0.25) = (-0.765, 0.25) and -0.765 is not between 0 and 2.

Are you sure you wrote the question down right? Perhaps you should give us the question exactly as you were given?

Answer 2

Your first point there appears to only have x and y and no z. I'm sorry. I thought the second point had x,y and z. Have trouble with "." and "," on my computer.

I have also corrected it for your mitake in the x-coordinate

Let us assume it is a quadratic type function.

y = ax^2 + bx

We have to find a and b.

0.5 = a(2^2) + b(2) = 4a + 2b ..... [EQ-1]

0.25 = a[4sin^2(-pi/8)] + b[-2sin(-pi/8)]

0.25/[sin(-pi/8)] = a[4sin(-pi/8)] - 2b ...... [EQ-2]

Add [EQ-1] and [EQ-2} to remove b:

0.5 + 0.25/[sin(-pi/8)] = a{4 + 4[sin(-pi/8)]}

a = (0.25/4){2 + 1/[sin(-pi/8)]} / {1 + [sin(-pi/8)]}

a = -0.06208

Use the point (2,0.5) to find b:

0.5 = (-0.06208)2^2 + b2

b = [0.5 + 4(0.06208)]/2

b = 0.374152

y = (-0.06208)x^2 + (0.374152)x

Check all points:

It goes through (0,0)

(2,0.5) -> y = (-0.06208)4 - (0.374152)2 = 0.5 this is right

(-2sin(-pi/8),0.25) = (0.76537,0.25)

y = (-0.06208)(0.76537)^2 + (0.374152)(0.76537)

y = -0.03636 + 0.286363

y = 0.25

So the line goes through all three points.

The line is: y = (-0.06208)x^2 + (0.374152)x