To solve the quadratic equation, \(\displaystyle \small 13= x^{2} -12x\), we set the equation equal to zero and then factor the quadratic, \(\displaystyle \small (x-13)(x+1)=0\). Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations \(\displaystyle x-13=0\) and     \(\displaystyle x+1=0\) to obtain the answers \(\displaystyle x=13\) and \(\displaystyle x=-1\). 

To factor this equation, first find two numbers that multiply to 35 and sum to 12.  These numbers are 5 and 7.  Split up 12x using these two coefficients:

\(\displaystyle x^2 + 7x+5x+35=0\)

\(\displaystyle x(x+7)+5(x+7)=0\)

\(\displaystyle (x+5)(x+7)=0\)

 \(\displaystyle x=-5,-7\)

To find \(\displaystyle x\), we must factor the quadratic function:

\(\displaystyle 3x^{2}-18x+24=0\)

\(\displaystyle 3(x^{2}-6x+8)=0\)

\(\displaystyle 3(x-4)(x-2)=0\)

\(\displaystyle x-4=0\, and \, x-2=0\)

\(\displaystyle x=4\,and\,2\)

To find \(\displaystyle x\), we want to factor the quadratic function:

\(\displaystyle 2x^{2}-10x-28=0\)

\(\displaystyle 2(x^{2}-5x-14)=0\)

\(\displaystyle 2(x-7)(x+2)=0\)

\(\displaystyle x-7=0\,and\,x+2=0\)

\(\displaystyle x=7\:and\:-2\)

Only equation B maps each value of \(\displaystyle x\) into a unique value of \(\displaystyle y\) and in a similar way each and every value of \(\displaystyle y\) maps into one and only one value of \(\displaystyle x\).

This question demonstrates that complicated functions are not complicated at every point.

To solve the function at x=1, all that is necessary is familiarity with the operations used.

\(\displaystyle f(1) = \frac{1^{4}-1+\sqrt[3]{1}}{1^{1}+3(1)}\)

\(\displaystyle = \frac{1-1+1}{1+3}\)

\(\displaystyle =\frac{1}{4}\)

To evaluate \(\displaystyle f(6)\) substitute six in for every x in the equation.

\(\displaystyle f(x) = x^{2} - 4x + 7\)

\(\displaystyle \\f(6) = 6^{2} - 4 \cdot 6 + 7 \\f(6)= 36 - 24 + 7\\f(6) = 12 + 7 \\f(6)= 19\)

To solve this problem replace every x in \(\displaystyle g(x)\) with \(\displaystyle x+6\).

\(\displaystyle g(x )= 3x - 13\)

Therefore,

\(\displaystyle g(x+ 6 )= 3(x+ 6 ) - 13\)

\(\displaystyle = 3\cdot x+ 3 \cdot 6 - 13\)

\(\displaystyle = 3 x+ 18- 13\)

\(\displaystyle = 3 x+ 5\)

Each of the tables provided contains sets of ordered pairs. The input column represents the x-variables, and the output column represents the y-variables. We can tell if a set of ordered pairs represents a function when we match x-values to y-values. 

In order for a table to represents a function, there must be one and only one input for every output. This means that our correct answer will have all unique input values:

Functions cannot have more than one input value that is the same; thus, the following tables do not represent a function: 

Each of the tables provided contains sets of ordered pairs. The input column represents the x-variables, and the output column represents the y-variables. We can tell if a set of ordered pairs represents a function when we match x-values to y-values. 

In order for a table to represents a function, there must be one and only one input for every output. This means that our correct answer will have all unique input values:

Functions cannot have more than one input value that is the same; thus, the following tables do not represent a function: