%path = "maths/functions/graph" %kind = kinda["texts"] %level = 10
Instructions: If you change the values in a field and go to the next field, the old curve stays on the diagram to allow comparison with the new one. Click [-1-] to show only the last curve.
A function equation (like y=1/x), determines a dependence of the two variables x and y.
Often x is chosen as independent variable
and y as dependent variable.
Note: Not every function has a equation and not every function can be represented as a curve.
If y is given and x can be calculated, then one swaps the names x and y to get the inverse function, which graphically results by mirroring at the line y=x (due to the swapping).
Exercise: Above a=1, n=1, u=0, v=0. This is y=x.
You make a table with two columns, one for x and one for y.
Then you choose more values for x equally space over the x range and write them in the x column.
Then you calculate y for each such x value and write it in the row with the used x value.
You use the values from the value table and draw the according points into the coordinate system, then you connect the points. Here the computer does it, but you should be able to do it on a piece of paper as well.
Exercise: Draw the graph of y=1/x on a piece of paper. Take positive and negative values for x. What is with 0?
\(y=x^n\)
Exercise:
a=1, n=2, u=0, v=0
a=1, n=3, u=0, v=0.
What symmetries do these curves have?
An even exponent makes the curve symmetric with at the y Achse.
Try n=4,
n=6,...
An uneven exponent makes the curve point-symmetric at the origin.
Try n=3,
n=5,...
E.g. \(y=x^{1/2}\)
\((x^2)^{(1/2)}=x^{(2/2)}=x\). The exponent 1/2=0.5 is the inverse of "to the power 2".
This is traditionally written with the root symbol: \(\sqrt{x}=x^{0.5}\).
\(x^{0.1}=x^{1/10}\) is the 10th root of x and the inverse of \(x^10\).
With ℝ, roots are defined only for positive numbers, that's why the graph with n=0.5 shows only the right part. The same is true for all none-integer exponents.
Exercise:
a=1, n=0.5, u=0, v=0.
Press [-1-].
Go to the n field, enter 2 and click on another field, but not on [-1-].
Go back to the n field and enter 1, then another field, but not [-1-].
Compare the two curves with y=x.
\(y=x^n\), n negative.
\(y=x^{-2}=1/x^2\), i.e. negative exponent yields the reciprocal. Where \(x^n\) is large, \(x^{-n}\) becomes small and vice versa.
Symmetry for negative n is the same as for positive n.
With x=0, one cannot compute \(1/x^n\), because it becomes infinite. This is called a Pole.
Exercise:
a=1, n=1, u=0, v=0
a=1, n=-1, u=0, v=0
Compare.
a=1, n=2, u=0, v=0
a=1, n=-2, u=0, v=0
Compare.
a=1, n=3, u=0, v=0
a=1, n=-3, u=0, v=0
Compare.
\(y=a(x-u)^n + v\)
Exercise:
Starting from a=1, n=2, u=0, v=0
set u to
-2,
-1,
0,
1,
2,...
set v to
-2,
-1,
0,
1,
2,...
Translation via u, v. If you replace x by x-u, then the graph moves u to the right, or to the left, if u is negative.
Set u = 2, then u = -2. With u=-2 we have x-(-2) = x+2.
Stretching and crushing and turn upside down with a.
Start from a=
1,
If a becomes larger (like a=
2),
then the curve becomes stretched.
If a=
0.5
or becomes even smaller, then the curve gets crushed.
With a=
-1,
the parabola (n=2) is turned upside down and opens downward.
\(y=b^x\)
e.g. \(y=2^x\)
Exercise:
For b=2 calculate in your head some function values for b=2 and x=-3,-2,-1,0,1,2,3
Note that \(b^0 = 1\) independent of the basis.
Above choose exponential function.
Set
a=1, b=2, u=0, v=0
Press [-1-]
Change b to
3,
4,
5,...
For b=1/2 calculate in your head some function values for e.g. x=-3,-2,-1,0,1,2,3
Then do above
a=1, b=0.5, u=0, v=0
Compare with basis b=2
The same difference as between b=2 und b=0.5 you get by replacing x and -x.
Try negative a values and change u and v.
The exponential function \(y=b^x\) is defined in all of ℝ.
The logarithm is the exponent. So \(log_2\) of 8 is "2's exponent to get 8". Or:
2 to the power 2 Logarithm of 8 is 8.
2 to the power 2 Logarithm of 16 is 16.
...
\(b^{\log_b(z)}=z\)
Because the exponent gets added when doing multiplication, we have:
\(\log{ab}=\log(a)+\log(b)\)
\(\log{\frac{a}{b}}=\log(a)-\log(b)\)
Similarly because of \((u^v)^w=u^{vw}\) we have
\(\log_b(c^d)=d\cdot\log_b(c)\) (\(=\log_b(b^{\log_b(c)\cdot d})\))
Exercise:
Calculate in your head the logarithm to the basis 2 of 1/32,1/16,1/8,1/4,1/2,0,2,4,8,16,32,...
The logarithm is the inverse of the exponential function.
Der graph therefore follows from the exponential function through mirroring at y=x.
Exercise:
Above choose logarithm with b=2, u=0, v=0
Compare the graph with the values calucated in your head.
Note: log 1 = 0, independent of which basis is used.
Change the basis. It must not be 1 though, why?:
0.25,
0.5,
0.75,
3,
5,...