The vertical line test can be used to determine if an equation is a function. In order to be a function, there must only be one \(\displaystyle \small y\) (or \(\displaystyle \small f(x)\)) value for each value of \(\displaystyle \small x\). The vertical line test determines how many \(\displaystyle \small y\) (or \(\displaystyle \small f(x)\)) values are present for each value of \(\displaystyle \small x\). If a single vertical line passes through the graph of an equation more than once, it is not a function. If it passes through exactly once or not at all, then the equation is a function.

The horizontal line test can be used to determine if a function is one-to-one, that is, if only one \(\displaystyle \small x\) value exists for each \(\displaystyle \small y\) (or \(\displaystyle \small f(x)\)) value. Calculating zeroes, domain, and range can be useful for graphing an equation, but they do not tell if it is a function.

Example of a function: \(\displaystyle y=x^2-6x\)

Example of an equation that is not a function: \(\displaystyle y^2-2x+y=7\)

THe notation \(\displaystyle f(g(x))\) is a composite function, which means we put the inside function g(x) into the outside function f(x). Essentially, we look at the original expression for f(x) and replace each x with the value of g(x).

The original expression for f(x) is \(\displaystyle 2-2x+x^2\). We will take each x and substitute in the value of g(x), which is 2x-1.

\(\displaystyle f(g(x))=f(2x-1)\)

\(\displaystyle =2-2(2x-1)+(2x-1)^2\)

\(\displaystyle =2-2(2x-1)+(2x-1)^2\)

We will now distribute the -2 to the 2x - 1.

\(\displaystyle 2-4x+2+(2x-1)^2\)

We must FOIL the \(\displaystyle (2x-1)^2\) term, because \(\displaystyle (2x-1)^2=(2x-1)(2x-1)\). 

\(\displaystyle 2-4x+2+(2x-1)^2=2-4x+2+(2x-1)(2x-1)\)

\(\displaystyle =2-4x+2+4x^2-2x-2x+1\)

Now we collect like terms. Combine the terms with just an x.

\(\displaystyle 2+2+4x^2+1-8x\)

Combine constants.

\(\displaystyle 4x^2-8x+5\)

The answer is \(\displaystyle 4x^2-8x+5\).

\(\displaystyle g(f(x))\) means \(\displaystyle f(x)\) gets plugged into \(\displaystyle g(x)\).

Thus \(\displaystyle g(f(x))= (3x^{2}+5) - 2 = 3x^{2}+3\).

Calculate \(\displaystyle g(2)\) and plug it into \(\displaystyle f(x)\).

\(\displaystyle g(2)= (2)^{2}-1=4-1=3\)

\(\displaystyle f(3)=(3)^{3}-2(3)^{2}+(3)=27-2(9)+(3)=27-18+3=12\)

\(\displaystyle \small f(g(3))\)

This expression is the same as saying "take the answer of \(\displaystyle \small g(3)\) and plug it into \(\displaystyle \small f(x)\)."

First, we need to find \(\displaystyle \small g(3)\). We do this by plugging \(\displaystyle \small 3\) in for \(\displaystyle \small x\) in \(\displaystyle \small g(x)\).

\(\displaystyle g(x)=x^2\)

\(\displaystyle g(3)=3^2=9\)

Now we take this answer and plug it into \(\displaystyle \small f(x)\).

\(\displaystyle f(g(3))=f(9)\)

We can find the value of \(\displaystyle \small f(9)\) by replacing \(\displaystyle \small x\) with \(\displaystyle \small 9\).

\(\displaystyle f(x)=6x-4\)

\(\displaystyle f(9)=6(9)-4=50\)

This is our final answer.