Mathematics: Algebraic Reasoning Study Guide for the TSIA2

Evaluating Functions

To evaluate a function for a specific value, substitute the value of the variable into the expression and then perform the indicated operations. The numeric result of the operations is the value of the function for the given value of the variable.

Basic Function Evaluation

Consider the following function: \(f(x) = 3x - 1\), which graphs as:

We can evaluate the function both algebraically and graphically. To evaluate, select an input value, \(x = 1\), for example, and substitute it into the function:

\[f(1) = 3(1) - 1 = 3 - 1 = 2\]

Graphically, this means that when \(x = 1\), the function evaluates to \(f(x) = 2\), which is the point \((1, 2)\). As can be seen from the graph, when \(x = 1\), \(f(x) = 2\).

Graphing Functions

Pairs of (x,y) values that are true for the function can be graphed with the cartesian coordinate system. The resulting line or curve is called the graph of the function.

Graphing Relations

Relations are graphed in the same way as functions. You may find that the graph is broken in places, or that there are large areas rather than lines, or that there is more than one value of y at any given value of x.

Quadratic Functions

A polynomial with one variable and a degree of 2 is called a quadratic, and functions equal to a quadratic are called quadratic functions. Quadratic functions can be written in the form:

\[y = ax^{2} + bx +c\]

General Procedure

Evaluating a quadratic function is exactly like evaluating any function: substitute and simplify. If you see \(f(2)\), for example, that tells you to substitute 2 in place of the variable in the function.

Example

If \(f(x)=x^2 -3x+11\), evaluate \(f(5)\).

\[f(5) = 5^2 -3\cdot 5 +11\] \[f(5) = 25-15+11\] \[f(5) = 21\]

Graphing Quadratic Functions

The graph of a quadratic function is a parabola. Parabolas resemble the letter U and can open upward, if \(a\) is positive, or downward, if \(a\) is negative:

Consider the following parabola with \(a > 0\), \(f(x) = 2x^2 - 2x + 1\), which graphs as:

Now consider the following parabola with \(a < 0\), \(f(x) = -2x^2 - 2x + 1\), which graphs as:

Notice that the positive parabola has its vertex as its lowest point, and that the negative parabola has its vertex as its highest point.

Vertex

The vertex is the highest or lowest point of the parabola, where it changes direction.

Discriminant

The quantity \((b^{2} - 4ac)\) is called the discriminant of the quadratic function. If the discriminant is negative, the parabola does not cross the x-axis and there are no solutions (also called roots or zeroes) that are real numbers.
If the discriminant is zero, the vertex of the parabola touches the x-axis and there is one real solution.
If the discriminant is positive, the parabola crosses the x-axis in two places and there are two real solutions.

Polynomial Functions

If a function \(f(x)\) is equal to a polynomial it is called a polynomial function.

General Procedure

Terms with the same variable(s) raised to the same power are called like terms. When adding or subtracting polynomials, only the like terms can be combined. The result of the operation will be a new, possibly simpler, polynomial.

When multiplying polynomials, all the terms in one polynomial must be multiplied against all the terms in the second one, and then like terms are combined. When both polynomials to be multiplied have two terms, there is a mnemonic called FOIL to remember the process:

Multiply First terms, then Outer terms, then Inner terms, then Last terms.

Example

If \(f(x) = 3(x^2-2x+1) +2(5x^2 -4)\), evaluate f(-2)

You can choose to substitute first or simplify first. In this case, we will simplify first.

Simplify: \(3(x^2-2x+1) +2(5x^2 -4)\)

\[3x^2 -6x +3 +10x^2-8\] \[13x^2 -6x-5\]

Substitute \(-2\) for \(x\).

\[13(-2)^2 -6(-2) -5\] \[13 \cdot 4 +12-5\] \[52+12\] \[64\]

Exponential Functions

Exponential functions are very similar to exponential equations, in that they involve quantities that have variables for exponents. For example, \(f(x)= 6^x -4\).

General Procedure

To evaluate an exponential function is just like evaluating any function or expression: substitute and simplify. That is, you first substitute the given value for the variable and then simplify the result. There are two reasons for evaluating a function. One is that a particular exercise directly asks you to do that. The second is to make a table of values for graphing the function.

Example

Suppose you wanted to graph the function \(f(x)= 4^x+1\). Complete the table:

\[\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2\\ \hline f(x) &? &? &? &? &? \\ \hline \end{array}\]

Substitute \(-2\) for \(x\) and simplify:

\[f(-2)= 4^{-2}+1\] \[f(-2)= \dfrac{1}{16}+1\] \[f(-2)= 1\dfrac{1}{16}\]

Substitute \(-1\) for \(x\) and simplify:

\[f(-1)= 4^{-1}+1\] \[f(-1)= \dfrac{1}{4}+1\] \[f(-1)= 1\dfrac{1}{4}\]

Substitute \(0\) for \(x\) and simplify:

\[f(0)= 4^0+1\] \[f(0)= 1+1\] \[f(0)= 2\]

Substitute \(1\) for \(x\) and simplify:

\[f(1)= 4^1+1\] \[f(1)= 4+1\] \[f(1)= 5\]

Substitute \(2\) for \(x\) and simplify:

\[f(2)= 4^2+1\] \[f(2)= 16+1\] \[f(2)= 17\]

To make the graph, you could now plot these points:

\[\begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2\\ \hline f(x) &1\frac{1}{16} &1\frac{1}{4} &2 &5 &17 \\ \hline \end{array}\]

The graph would look like this, though the point \((2, 17)\) isn’t shown.

Rational Functions

Rational functions are similar to rational expressions, except that the fraction is set equal to a second variable. Since there is a denominator, there exists the possibility that the denominator is zero for some value of x. In that case, the function is not defined for that value of x and there is a break in the graph at that point.

General Procedure

Evaluating rational functions is no different than evaluating any other function: Substitute and simplify. However, as mentioned above, you do have to worry about any value that will make the denominator zero, as the function will be undefined for that value.

Example

The function \(y = \frac{2x + 1}{x-2}\) is undefined at \(x = 2\), because that value of \(x\) results in a denominator of zero. Its graph would look like this:

Notice that the function gradually approaches \(x = 2\), but never touches it; \(x = 2\) is a vertical asymptote.

Radical Functions

A radical function is any function that is defined by a radical expression.

For example:

\[\begin{array}{ccc} f(x) = \sqrt{3x-4} &\quad g(x) = \sqrt[3]{x^2+7} & \quad h(x) = \sqrt{-5x^2-17}\\ \end{array}\]

General Procedure

Radical functions are evaluated with the same substitute and simplify procedure as other functions, but you do need to be aware that some inputs will produce values that are not real numbers. This happens when you have a square root (or fourth root, or sixth root, etc.) of a negative number.

Example 1

If \(f(x) = \sqrt{3x+1}\), evaluate \(f(5)\)

\[f(5) = \sqrt{3 \cdot 5 +1}\] \[f(5) = \sqrt{16}\] \[f(5) = 4\]

Example 2

If \(f(x) = \sqrt{8-2x^2}\), evaluate \(f(3)\)

\[f(3) = \sqrt{8-2 \cdot 3^2}\] \[f(3) = \sqrt{8-18}\] \[f(3) = \sqrt{-10}\]

This gives us the square root of a negative number, which is not a real number. You may recognize \(\sqrt{-10}\) as the imaginary number \(i\sqrt{10}\).