class 051007

What are the basic points of the basic linear, quadratic and power functions?

 

The first graph below represents a linear function, the second represents a quadratic function and the third represent a p = -1 power function.

 

basic points

 

 

 

 

What are the basic points for y=f(x)?

The answer depends on the function f(x).

The basic linear function is y = x and its basic points are the x = 0 and x = 1 points.

The basic quadratic function is y = x^2 and its basic points are the x = -1, x = 0 and x = 1 points.

The basic points of the exponential function y = 2^x are the x = 0 and x = 1 points, and also the horizontal asymptote at the negative x axis. The asymptote isn't really a point but we include it among the three basic things we need in order to quickly graph an exponential function.

The basic points of a power function y = x^p are the x = -1, x = 0, x = 1/2, x = 1 and x = 2 points. If p is negative, then the x = 0 'point' isn't really a point but a vertical asymptote, and as the trend of basic points shows, for negative p the x axis is a horizontal asymptote in both directions.

If a function is of the form y = A f(x-h) + k then the basic points (and all points) are 'stretched' to positions A times as far from the x axis, horizontally shifted h units and vertically shifted k units. By applying this principle just to the basic points, we can quickly and easily locate the graph of A f(x-h) + k for any of our basic functions.

The basic points of y = A f(x-h) + k are the points you get when you apply these stretches and shifts to the basic points of the function f(x).

Difference Equations

f(0) = 2

f(1) = 5

f(2) = 8

f(3) = 11

etc.

The pattern here goes up by 3 and starts at 2.  In more precise language we say

f(0) = 2

f(n+1) = f(n) + 3.

To use this, we first let n = 0.

Substituting 0 for n into f(n+1) = f(n) + 3 we get

f(0+1) = f(0) + 3

f(1) = 2 + 3

f(1) = 5.

Next, we let n = 1.  Substituting 1 for n into f(n+1) = f(n) + 3 we get

f(1 + 1) = f(1) + 3

f(2) = f(1) + 3.  Since f(1) = 5 from the previous step:

f(2) = 5 + 3

f(2) = 8.

Next, we let n = 2.  Substituting 2 for n into f(n+1) = f(n) + 3 we get

f(2 + 1) = f(2) + 3

f(3) = f(2) + 3.  Since f(2) = 8 from the previous step:

f(3) = 11.

It's clear that this process just keeps adding 3 to the previous value to get the next value.

What linear function f(x) would give us the same values?

A graph of f(n) vs. n goes up 3 for every unit of run so its slope is 3.

The graph goes through (0, 2) so the y intercept is 2.

The function is therefore y = m x + b with m = 3 and b = 2. 

The function is y = 3x + 2.

The linear function y = f(x) = m x + b can be modeled by the equations

f(0) = b

f(n+1) = f(n) + m.