Equation of function

**SHiFT** · October 21st 2009, 11:36 AM

f is a linear function such that f(1) = 0 and f(2) = 4. How do you write an equation for f(x)

**artvandalay11** · October 21st 2009, 11:39 AM

Originally Posted by SHiFT

f is a linear function such that f(1) = 0 and f(2) = 4. How do you write an equation for f(x)

I'm sure you heard somewhere along the line that linear functions can be written as $y=mx+b$ or $f(x)=mx+b$

So we have 2 points, and that's all we need. We use them to plug into that general formula

$0=m(1)+b$ and $4=m(2)+b$

Solve those two equations for m and b and you'll be able to write down your line $y=mx+b$

**SHiFT** · October 21st 2009, 11:43 AM

I'm still confused though, so do I need to come up with two of my own points then solve, or do I just solve with the variables given?

**artvandalay11** · October 21st 2009, 11:50 AM

Originally Posted by SHiFT

I'm still confused though, so do I need to come up with two of my own points then solve, or do I just solve with the variables given?

You need to solve the following system of equations

1. 0=m(1)+b and

2. 4=m(2)+b

Now let's try to get you to understand this. I hand you two points, (p,q) and (r,s) and say what's the equation of the line that goes through them

You realize that all lines on the x-y plane can be written as $y=mx+b$

So you also know that for the ordered pair (p,q), the x coordinate is p and the y coordinate is q.... same for (r,s)

So since $y=mx+b$ must be satisfied, you plug your points into in to get

q=mp+b and

s=mr+b

Now in a normal problem p,q,r,s will be numbers so the only "unknowns" here are m and b

You have two equations and 2 unknowns so you can solve for both m and b, and then the line you were looking for is y=mx+b

Here they tell you f(1)=0 and f(2)=4, so since f(x)=y, the points they give you are (1,0) and (2,4)

**masters** · October 21st 2009, 12:08 PM

Originally Posted by SHiFT

f is a linear function such that f(1) = 0 and f(2) = 4. How do you write an equation for f(x)

Hi SHiFT,

If f(1)=0, then the point is (1, 0).

If f(2)=4, then the point is (2, 4)

Find the slope using your two points.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{2-1}=4$

Now, plug your slope of 4 and point (1, 0) into $y=mx+b$ and find y-intercept b.

$0=4(1)+b$

$b=-4$

$y=4x-4$

$\boxed{f(x)=4x-4}$

**artvandalay11** · October 21st 2009, 12:18 PM

Originally Posted by masters

Hi SHiFT,

If f(1)=0, then the point is (0, 1). This is your y-intercept.

If f(2)=4, then the point is (4, 2)

Find the slope using your two points.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-1}{4-0}=\frac{1}{4}$

Now, plug your slope and y-intecept into $y=mx+b$

$y=\frac{1}{4}x+1$

$\boxed{f(x)=\frac{1}{4}x+1}$

We have 2 conditions, f(1)=0 and f(2)=4

You gave $\boxed{f(x)=\frac{1}{4}x+1}$

And if we compute f(1) we get $\frac{1}{4}(1)+1=1.25\not = 0$

And if we compute f(2) we get $\frac{1}{4}(2)+1=1.5\not=4$

You swapped your x and y's, meaning this function takes 0 to 1 and 4 to 2, you gave us the inverse of the function is question

**masters** · October 21st 2009, 12:23 PM

Originally Posted by artvandalay11

We have 2 conditions, f(1)=0 and f(2)=4

You gave $\boxed{f(x)=\frac{1}{4}x+1}$

And if we compute f(1) we get $\frac{1}{4}(1)+1=1.25\not = 0$

And if we compute f(2) we get $\frac{1}{4}(2)+1=1.5\not=4$

You swapped your x and y's, meaning this function takes 0 to 1 and 4 to 2, you gave us the inverse of the function is question

Excellent observation. I was just testing you.....I shall fix it straight away.

**SHiFT** · October 21st 2009, 12:49 PM

Thanks a ton guys, it makes a lot of sense now

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