Quantitative Reasoning Study Guide for the DAT

Algebra

Algebra serves as the language of mathematics, providing a systematic approach to representing and solving mathematical problems using symbols and operations. It encompasses a wide range of concepts, including equations, expressions, variables, and functions. Algebraic techniques allow for the solution of complex problems, modeling of real-world phenomena, and formulation of mathematical relationships.

The Vocabulary of Algebra

There are some common terms in algebra that you should be familiar with:

• constant—A constant is a fixed numerical value that does not change throughout a problem or equation. Constants can be integers, fractions, decimals, or irrational numbers. In algebraic expressions and equations, constants are represented by specific numerical values. For example, in the expression \(4x+2\), the constants are \(4\) and \(2\).

• variable—A variable is a symbol that represents an unknown or changing quantity in algebraic expressions and equations. Variables are often denoted by letters such as \(x\), \(y\), or \(z\), and their values can vary depending on the context of the problem. For example, in the equation \(3x - 7 = 10\), \(x\) is the variable representing the unknown quantity.

• expression—An expression is a mathematical phrase that combines numbers, variables, and operations such as addition, subtraction, multiplication, division, and exponentiation. Expressions can be simple or complex and may involve multiple terms and operations. Examples of expressions include \(2x+5\), \(3y^{2} - 4y + 7\), and \(\frac{a}{b} + c\).

• equation—An equation is a mathematical statement that asserts the equality of two expressions. Equations involve two sides connected by an equal sign (\(=\)). Solving equations involves finding values of variables that make the equation true. For example, the equation \(2x - 3 = 7\) asserts that the expression \(2x - 3\) is equal to \(7\).

• term —A term is a component of an algebraic expression separated by addition or subtraction. A term can be a constant, a variable, or a product of constants and variables. Terms are essential for understanding the structure of expressions and equations and play a crucial role in algebraic manipulation. For example, in the expression \(3x^{2} + 5xy -2\), \(3x^{2}\), \(5xy\), and \(-2\) are the individual terms.

Solving Word Problems Using Algebra

Word problems in algebra require the translation of verbal descriptions into mathematical expressions or equations. The key to successfully solving these problems lies in identifying and understanding the key words and phrases that indicate specific mathematical operations and relationships between quantities. By recognizing these key words, we can break down the problem into meaningful components and assign variables to represent the unknown quantities.

For instance, consider the phrase “three more than a number.” Here, we can assign a variable, such as \(x\), to represent the unknown number. The phrase “three more than” indicates addition, so the corresponding mathematical expression would be \(x+3\). Similarly, let’s take the phrase “twice a number decreased by five.” Again, we assign a variable, \(y\), to represent the unknown number. The phrase “twice” indicates multiplication by two, and “decreased by” indicates subtraction, resulting in the mathematical expression \(2y- 5\). By systematically translating verbal statements into algebraic expressions or equations using these key phrases, we can effectively solve word problems and interpret the results within the context of the problem.

The table below shows common key words and their corresponding mathematical symbols:

English	Mathematical symbol
plus, sum, combined, added to	\(+\)
subtracted from, minus, smaller than, difference	\(-\)
multiplied by, times, of, product of	\(\times\)
divided by, out of, per, ratio	\(\div\)
equals, is, will be, was, has	\(=\)

Exponents

Exponents are mathematical notations that represent the repeated multiplication of a number by itself. They are used to express powers and describe the growth or decay of quantities over time. Exponents play a crucial role in algebra by simplifying expressions, solving equations, and modeling exponential relationships. Understanding exponent rules allows for the efficient manipulation and evaluation of algebraic expressions involving powers.

Remembering the following rules for exponents will help you manipulate and solve exponentials:

• product rule—\(x^m \times x^n = x^{m+n}\)

• quotient rule—\(\frac{x^m}{x^n} = x^{m-n}\)

• power rule—\((x^m)^{n} = x^{m \times n}\)

• zero exponent rule—\(x^{0} = 1\)

• negative exponent rule—\(x^{-n} = \frac{1}{x^n}\)

Let’s look at an example.

Simplify: \(\frac{y^6 \times y^{-3}}{y^4}\)

Solution

First, we use the product rule to simplify the numerator:

\[\frac{y^6 \times y^{-3}}{y^4}\] \[= \frac{y^{6 + (-3)}}{y^4}\] \[= \frac{y^3}{y^4}\]

Now, we use the quotient rule for exponents:

\[= y^{3-4}\] \[= y^{-1}\]

This is the answer. To simplify further, we can use the negative exponent rule to write:

\[y^{-1} = \frac{1}{y^1} = \frac{1}{y}\]

Logarithms

Logarithms are functions that offer powerful tools for solving exponential equations, understanding exponential growth and decay, and simplifying calculations involving large or small quantities. Logarithms provide a systematic way to express exponential relationships in a more manageable form. They allow us to convert multiplication into addition and exponentiation into multiplication, making complex calculations more accessible. In essence, logarithms offer a means to “shrink” or condense large or small numbers into a more convenient range, facilitating easier computation and analysis.

Basics

The logarithm of a number, \(x\), to a given base, \(b\), is denoted as \(\log_{b}(x)\). It represents the exponent to which the base \(b\) must be raised to produce the number \(x\). In other words, if \(b^y = x\), then \(y = \log_{b}(x)\). This is the conversion form of logarithms to exponentials.

Properties of Logarithms

Here, we will look at three important properties of logarithms that help us manipulate and solve logarithmic and exponential equations and expressions:

• product property—This property states that the logarithm of the product of two numbers is equal to the sum of their logarithms: \(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\). This allows us to break down complex multiplications into simpler additions, simplifying calculations. For example, we can write \(\log_{10}(50)\) as \(\log_{10}(5 \times 10) = \log_{10}(5) + \log_{10}(10)\).

• quotient property—Similar to the product property, the quotient property states that the logarithm of the quotient of two numbers is equal to the difference of their logarithms: \(\log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y)\). This property helps simplify divisions by converting them into subtractions. For example, we can write \(\log_{10}(40)\) as \(\log_{10}\left(\frac{80}{2}\right) = \log_{10}(80) - \log_{10}(2)\).

• power property—The power property of logarithms states that the logarithm of a number raised to a power is equal to the product of the exponent and the logarithm of the base. In equation form, \(\log_{b}(x^n) = n \log_{b}(x)\). This allows us to simplify exponentiations into multiplications, facilitating calculations. Using this property, we can write \(\log_{3}(25)\) as \(\log_{3}(5^2) = 2 \log_{3}(5)\).

Polynomials

Polynomials are algebraic expressions consisting of variables and coefficients. They represent numbers that are the result of operations, like addition, subtraction, multiplication, and division. One example of a polynomial is:

\[P(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + … + a_{1} x + a_{0}\]

where \(a_n\), \(a_{n-1}\), \(a_1\), and \(a_0\) are constants, \(x\) is a variable, and \(n\) is a non-negative integer representing the degree of the polynomial.

There are several parts of a polynomial:

• terms—Each part of a polynomial that is separated by an addition or subtraction sign is called a term. For example, in the polynomial \(3x^2 - 5x + 2\), \(3x^2\), \(5x\), and \(2\) are the terms.

• coefficients—Coefficients are the numerical factors that multiply the variable in each term. In the term \(3x^2\), the coefficient is \(3\).

• exponents— Exponents represent the power to which the variable is raised. In the term \(3x^2\), the exponent of \(x\) is \(2\).

You should be familiar with the following types of polynomials:

A monomial is a polynomial with only one term, for example, \(5x\) or \(-2y^{3}\).

A binomial is a polynomial with two terms, for example, \(2a+b\).

A trinomial is a polynomial with three terms, for example, \(2y^{3} + 7y + 2\).

Operations with Polynomials

We can perform the same four operations with polynomials as we do with numbers.

Addition of Polynomials

To add polynomials, simply combine like terms, which are terms that have the same variable raised to the same exponent. For example:

\[(3x^2 + 2x - 4) + (x^2 - 5x + 6)\] \[= (3x^2 + x^2) + (2x - 5x) + (-4 + 6)\] \[= 4x^2 - 3x + 2\]

Subtraction of Polynomials

Subtracting polynomials is similar to adding them. Distribute the negative sign and then combine like terms. For example:

\[(3x^2 + 2x - 4) - (x^2 - 5x + 6)\] \[= 3x^2 + 2x -4 -x^2 + 5x - 6\] \[= (3x^2 - x^2) + (2x + 5x) + (-4-6)\] \[= 2x^2 + 7x - 10\]

Multiplication of Polynomials

Multiplying polynomials involves distributing each term of one polynomial to every term of the other polynomial and then combining like terms. It applies to all types (e.g., monomial, binomial, trinomial) of polynomials. Here is an example of multiplying a binomial by another binomial:

\[(2x + 2)(-4x + 6)\] \[= 2x(-4x) + 2x(6) + 2(-4x) + 2(6)\] \[= -8x^2 + 12x - 8x + 12\] \[= -8x^2 + 4x + 12\]

Division of Polynomials

The division of polynomials can be done using long division, similar to the division of numbers. It involves dividing the highest-degree term of the dividend by the highest-degree term of the divisor and repeating the process until the degree of the remainder is less than the degree of the divisor.

If you are asked to obtain the remainder of a polynomial division, the remainder theorem can be helpful. The remainder theorem says that if a polynomial, \(p(x)\), is divided by another polynomial of the form \(x - a\), the remainder is simply \(p(a)\). It is important to note that this only applies when the polynomial divisor is of the form \(x - a\).

Let’s look at a more general example, where we divide the polynomial \(4x^3 -3x^2+2x -5\) by the binomial \(2x - 1\):

To perform this long division:

1. Divide the highest-degree term of the dividend by the highest-degree term of the divisor:

\[4x^3 \div 2x = 2x^2\]

1. Multiply the entire divisor by the result obtained in step \(1\) and subtract it from the dividend:

\[2x^2 \times (2x - 1) = 4x^3 - 2x^2\] \[(4x^3 -3x^2+2x -5) - (4x^3 - 2x^2) = -x^2 + 2x - 5\]

1. Repeat the process until the degree of the remainder is less than the degree of the divisor:

\[-x^2 \div 2x = -\frac{1}{2}x\] \[-\frac{1}{2}x \times (2x - 1) = -x^2 + \frac{1}{2}x\] \[(-x^2 + 2x - 5) - (-x^2 + \frac{1}{2}x) = \frac{3}{2}x - 5\] \[\frac{3}{2}x \div 2x = \frac{3}{4}\] \[\frac{3}{4} \times (2x - 1) = \frac{3}{2}x - \frac{3}{4}\] \[(\frac{3}{2}x - 5) - (\frac{3}{2}x - \frac{3}{4}) = -\frac{17}{4}\]

Since the degree of the remainder (\(0\)) is less than the degree of the divisor (\(1\)), we are done and are left with:

\[4x^3 -3x^2+2x -5 \div 2x - 1 = 2x^2 - \frac{1}{2}x + \frac{3}{4}\]

with a remainder of \(-\frac{17}{4}\)

Factoring Polynomials

There are many methods for factoring polynomials. Here, we will look at the three most common methods.

Factor Common Terms

When factoring polynomials, it’s often helpful to first identify if there’s a common factor present in all terms. A common factor is a term that divides evenly into each term of the polynomial. To factor out the greatest common factor (GCF), follow these steps:

1. Identify the GCF.—Find the largest term that divides evenly into each term of the polynomial.

2. Factor out the GCF.—Divide each term by the GCF and rewrite the polynomial.

Consider the binomial \(6x^2 + 12x\).

First, identify the GCF. In this case, that is \(6x\). Now, factor out the \(6x\) from all the terms of the polynomial:

\[6x^2 + 12x = 6x(x+2)\]

Difference of Squares

The difference of squares is a special case where a polynomial can be factored into the product of two binomials. The formula for factoring the difference of squares is:

\[a^2 - b^2 = (a+b)(a-b)\]

Using this formula, you can easily factor the polynomial \(x^2 - 9\).

\(x^2 - 9\) can be written as \((x)^{2} - (3)^{2}\). Putting that into the formula, we have:

\[x^2 - 9 = (x+3)(x-3)\]

Multiplying Binomials

Multiplying a pair of binomials can be done easily with the FOIL method, which stands for “first, outside, inside, last.” Given a pair of binomials, to expand them into one polynomial expression, you need to multiply the first terms of both binomials, then multiply the “outside” terms of both binomials, then multiply the “inside” terms of both binomials, and finally, multiply the last terms of both binomials. Then, add all of these products together.

For example, if we want to multiply \((5x + 2)\) by \((3x - 5)\), we can foil:

First: \((5x)(3x) = 15x^2\)

Outside: \((5x)(-5) = -25x\)

Inside: \((2)(3x) = 6x\)

Last: \((2)(-5) = -10\)

Now, we add all these together and combine like terms:

\[(15x^2) + (-25x) + (6x) + (-10) = 15x^2 - 19x - 10\]

Square of Binomials

The square of a binomial is another special case where a polynomial expression can be factored into the product of two identical binomials. The formulas for factoring the square of binomials are:

\[a^2 + 2ab + b^2 = (a+b)^2\]

and

\[a^2 - 2ab + b^2 = (a-b)^2\]

For example, the polynomial \(x^2 -4x + 4\) can be factored as \((x-2)^{2}\) using the square of binomials formula. This is basically the opposite of the FOIL method.

Solving Equations

Solving equations involves finding the values of the variables in that equation. Equations are used to represent relationships between quantities and are used in various fields such as physics, engineering, and finance. Two of the most common types of equations we need to know how to solve are linear and quadratic. You should also be familiar with how to solve a set of simultaneous equations.

Linear Equations

Linear equations are equations of the form \(ax + b = c\), where \(x\) is the variable, and \(a\), \(b\), and \(c\) are constants. The main goal when solving linear equations is to isolate the variable \(x\) on one side of the equation.

The golden rule for solving linear equations is to perform the same operation on both sides of the equation to maintain its equality. This typically involves operations such as addition, subtraction, multiplication, and division.

Let’s consider the following linear equation:

\[5x + 10 = 20 + 3x\]

Our goal is to solve for the variable \(x\). First, let’s move all \(x\)s to one side by subtracting \(3x\) from both sides:

\[5x + 10 - 3x = 20 + 3x - 3x\] \[5x - 3x + 10 = 20\]

Now, we will move all the numbers to the other side by subtracting \(10\) from both sides:

\[5x - 3x + 10 - 10 = 20 - 10\] \[5x - 3x = 10\]

Now, we simplify to get:

\[2x = 10\]

To isolate \(x\), we divide both sides by \(2\):

\[\frac{2x}{2} = \frac{10}{2}\] \[x = 5\]

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(x\) is the variable and \(a\), \(b\), and \(c\) are constants. Quadratic equations have one variable to the degree of \(2\) and may have zero, one, or two real solutions.

There are three main methods for solving quadratic equations: factoring, completing the square, and using the quadratic formula. However, the quadratic formula is the most widely applicable method as it can be used to solve any quadratic equation.

Using the Quadratic Formula

The quadratic formula is given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Let’s use this formula to solve the quadratic equation \(x^2 + x - 6 = 0\).

To use the quadratic formula correctly, we need to know the values of \(a\), \(b\), and \(c\). Remember that the coefficient of \(x^2\) is \(a\), the coefficient of \(x\) is \(b\), and the constant term is \(c\). From the above equation, we see that \(a = 1\), \(b = 1\), and \(c = -6\). Make sure to take the sign of the constant also.

Now, using the quadratic formula, we have:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] \[x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-6)}}{2(1)}\] \[x = \frac{-1 \pm \sqrt{25}}{2}\] \[x = \frac{-1 \pm 5}{2}\] \[x = \frac{-1+5}{2} , \frac{-1-5}{2}\] \[x = \frac{4}{2} , \frac{-6}{2}\] \[x = 2, -3\]

This quadratic equation has two distinct real solutions. To determine the number of real solutions to a quadratic equation, we look at the term \(\sqrt{b^2 - 4ac}\) in the quadratic formula. This term is typically called the discriminant. When the discriminant is greater than zero, there are two (distinct) real solutions. When the discriminant equals zero, there is one real solution (in other words, two real solutions that are the same). When the discriminant is less than zero, there are no real solutions.

Factoring

Factoring is another method used to solve quadratic equations. In this method, we express the quadratic equation in the form of two binomial factors. The steps to factor a quadratic equation involve finding two numbers that multiply to give the constant term \(c\) and add to give the coefficient of \(x\), \(b\). Let’s solve the quadratic equation \(x^2+5x+6=0\) by factoring.

We need to find two numbers that when multiplied give us \(6\) and when added give us \(5\). Two such numbers are \(2\) and \(3\). Thus, we can factor the equation as:

\[x^2 + 5x + 6 = 0\] \[x^2 + 2x + 3x + 6 = 0\] \[x(x+2) + 3(x+2) = 0\] \[(x+2)(x+3) = 0\]

Now, we set each factor equal to \(0\) and find each of the two solutions:

\[x + 2 = 0\] \[x = -2\]

and

\[x+ 3 = 0\] \[x = -3\]

Note: If there was a coefficient of \(x^2\), a prerequisite step would be to multiply the coefficient of \(x^2\) by the constant \(c\). This is the number that would be the product of the two numbers we ultimately need to find to factor the equation.

Completing the Square

Completing the square is a method used to solve quadratic equations by rewriting the quadratic expression as a perfect square trinomial. This method involves adding and subtracting a constant, \(\left(\frac{b}{2}\right)^{2}\) (where \(b\) is the coefficient of \(x\)), to the equation to make it a perfect square trinomial. Let’s solve the same equation \(x^2 + 5x + 6 = 0\) from above using this method.

For the equation given, \(a = 1\), \(b =5\), and \(c = 6\). To complete the square, first, we move the constant to the other side:

\[x^2 + 5x + 6 = 0\] \[x^2 + 5x = - 6\]

To complete the square, we need to add and subtract \(\left(\frac{b}{2}\right)^{2}\) (or add it to both sides) to the equation:

\[x^2 + 5x + \left(\frac{5}{2}\right)^{2} = -6 + \left(\frac{5}{2}\right)^{2}\]

We simplify the expression and rewrite the left side as a perfect square trinomial:

\[x^2 + 5x + \frac{25}{4} = \frac{1}{4}\] \[\left(x+\frac{5}{2}\right)^{2} = \frac{1}{4}\]

Take the square root of both sides and find the two values of \(x\):

\[x + \frac{5}{2} = \pm \frac{1}{2}\] \[x = \pm \frac{1}{2} - \frac{5}{2}\] \[x = -2, -3\]

As you see, we have the same solutions that we got using the quadratic formula.

Simultaneous Equations

Simultaneous equations, also known as systems of equations, are equations involving two or more variables that are solved together. The goal is to find the values of the variables that satisfy all the equations simultaneously.

There are two common methods for solving simultaneous equations: the elimination method and the substitution method.

Consider the simultaneous equations shown below:

\[\begin{cases} x + y = 6 \\ -3x + y = 2 \end{cases}\]

First, let’s solve this using the elimination method. In the elimination method, we manipulate the equations such that when added or subtracted, one variable is eliminated, allowing us to solve for the other variable.

Let’s multiply the first equation by \(-1\):

\[-1 \times (x+y = 6)\] \[-x - y = -6\]

Now, we will add both equations. This will eliminate \(y\), and we will solve for \(x\):

\[(-x - y = -6) + (-3x + y = 2)\] \[-4x = -4\]

Solving for \(x\), we get:

\[x = \frac{-4}{-4} = 1\]

Now, we substitute this value of \(x\) into one of the original equations and find the corresponding value of \(y\):

\[x + y = 6\] \[1 + y = 6\] \[y = 5\]

To solve this using the substitution method, we would solve for any variable (such as \(x\)) in the first equation and put that into the second equation. Then, we will solve for \(y\). Let’s do this:

\[x + y = 6\] \[x = 6 - y\]

Substituting this into the second equation and solving for \(y\), we get:

\[-3x + y = 2\] \[-3(6-y) + y = 2\] \[- 18 + 3y + y = 2\] \[4y = 2 + 18\] \[4y = 20\] \[y = 20 \div 4 = 5\]

Now, we can substitute this value of \(y\) into the first equation and solve for \(x\):

\[x + y = 6\] \[x + 5 = 6\] \[x = 1\]

No matter which method we use, we get the same answer.

Graphing Equations

Graphing equations is essential for understanding and analyzing relationships between variables. Proficiency in graphing allows you to interpret data, solve problems, and comprehend mathematical concepts visually.

Graphing equations involves plotting points on a coordinate plane to represent mathematical functions or relationships. By visualizing these equations, one can observe patterns, trends, and solutions that may not be immediately apparent from the equations themselves.

In this section, we’ll cover the common types of graphs you need to know, including:

• linear functions
• quadratic functions
• radical functions
• piecewise functions

Each type of function has unique characteristics and key points that are essential to understand for accurate graphing and graph interpretation.

Basic Graphing

Graphing any function starts with understanding the \(x\)-intercepts and \(y\)-intercepts.

• \(x\)-intercepts are points where the graph crosses the \(x\)-axis (\(y = 0\)).
• \(y\)-intercepts are points where the graph crosses the \(y\)-axis (\(x = 0\)).

To find the \(x\)-intercept of a function, you set \(y=0\) and solve for \(x\). To find the \(y\)-intercept, you set \(x=0\) and solve for \(y\).

Here are some common function types:

• Linear functions are represented by straight lines that can be horizontal, vertical, or sloped.

• Quadratic functions form parabolas, U-shaped figures that either open upward or downward.

• Radical functions are graphs that include a square root (or other roots), typically starting from a point and curving in one direction. Negative values are not allowed inside the radical.

• Piecewise functions consist of different segments of functions defined over specific intervals. Each piece can have a different rule.

Graphing Linear Equations

Linear equation graphs are lines of the form \(y=mx+b\), where \(m\) is the slope of the line and \(b\) is the \(y\)-intercept. The slope is the ratio of \(y\) to \(x\). Special cases for linear graphs are horizontal and vertical lines.

• horizontal line: \(y = b\), where \(b\) is a constant

• vertical line: \(x = a\), where \(a\) is a constant

Equation of lines can be in different forms, but when graphing, it’s best to convert them to the slope-intercept form (\(y=mx+b\)) and then graph on the coordinate plane. Let’s graph the linear equation \(y = 2x +3\). Follow the steps shown below:

1. Identify the slope (\(m = 2\)) and \(y\)-intercept (\(b = 3\)).

2. Plot the \(y\)-intercept \((0, 3)\).

3. Use the slope to find another point. From \((0, 3)\), move up two units and right one unit to \((1, 5)\). Since the slope is \(2\), we can write it as \(\frac{2}{1}\), which, as we’ve already seen, means two units up \((y)\) and one unit right \((x)\).

4. Draw a line through these points.

You could plot more than two points, but two is the minimum number of points needed to draw the graph of a linear function.

Graphing Quadratic Equations

Quadratic equations produce parabolas, which are U-shaped curves. These equations have the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The leading coefficient, \(a\), determines the direction and width of the parabola. If \(a > 0\), the parabola opens upward, and if \(a < 0\), it opens downward. Graphing a quadratic equation involves finding the vertex, axis of symmetry, and additional points to accurately sketch the parabola.

We’ll graph the parabola \(y = x^2 - 4x + 3\) to see the steps that are required.

First, find the vertex, either the lowest or highest point of the parabola. The \(x\)-coordinate of the vertex is found using the formula \(x = -\frac{b}{2a}\).

Then, we take the \(x\)-coordinate and plug it back into the original equation to get the corresponding \(y\)-coordinate.

From the equation, we see that \(a = 1\), \(b = -4\), and \(c = 3\). The \(x\)-coordinate of the vertex is:

\[x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2\]

The \(y\)-coordinate is \(y = (2)^2 - 4(2) + 3 = -1\). So, the vertex is at \((2,-1)\).

Note that the axis of symmetry, the imaginary line that divides a parabolic graph into two equal halves, has the equation \(x = -\frac{b}{2a}\). So the axis of symmetry is \(x = 2\).

Next, we find the \(y\)-intercept of the graph by setting \(x=0\).

\[y = (0)^{2} - 4(0) + 3 = 3\]

Thus, the \(y\)-intercept is at \((0,3)\).

Since \(a >0\), the parabola will open upward. We can find additional points by substituting values into \(x\) to make the graph more precise. The graph is shown below:

Graphing Functions

Graphing other functions involves understanding the domain and range of the function, as well as identifying specific characteristics of different function types. The domain of a function consists of all possible input values (\(x\)-values), while the range consists of all possible output values (\(y\)-values). These are crucial for determining where a function is defined and the possible values it can output. Let’s look at two other types of functions: radical and piecewise.

Radical Function Graphs

Radical functions involve roots and have specific domain restrictions to ensure the expression under the radical is non-negative. Graphing a radical function requires identifying the starting point (where the expression under the radical equals zero) and plotting additional points to sketch the curve. Radical functions typically take the form \(y = \sqrt{x-a} + b\). The domain is restricted to values where the expression under the radical is non-negative (i.e., \(x - a \geq 0\)).

You can follow these steps to draw a radical function:

1. Determine the domain.—Solve \(x-a \geq 0\) to find the smallest \(x\)-value.

2. Identify the starting point.—This is the point where the expression under the radical equals zero, typically at \(x=a\).

3. Plot additional points.—Choose \(x\)-values greater than \(a\) and calculate the corresponding \(y\)-values.

4. Draw the curve.—Start at the initial point and extend the curve smoothly, noting the direction it moves.

Let’s graph the radical function with the equation \(y = \sqrt{x-1}\).

First, we determine the domain:

\[x - 1 \geq 0\] \[x \geq 1\]

Then, let’s find our starting point. At \(x = 1\), \(y = \sqrt{1-1} = 0\). So, the point is at \((1,0)\).

Now, let’s find two other points. At \(x =2\), \(y = \sqrt{2-1} = 1\) and at \(x=5\), \(y = \sqrt{5-1} = 2\). The two points are \((2,1)\) and \((5,2)\).

We can now draw the curve:

Piecewise Function Graphs

Piecewise functions consist of multiple segments, each defined by a different equation over specific intervals. Graphing piecewise functions involves graphing each segment separately and ensuring continuity at the boundary points between segments. You can follow these steps to graph piecewise functions:

1. Identify each piece and its interval.—Note the domain for each part of the function.

2. Graph each piece individually.—Plot points and draw the graph for each interval.

3. Check for continuity.—Determine if there are any jumps or breaks at the boundaries of the intervals. A discontinuity is a point where the graph of a function doesn’t continue.

Let’s graph the function with the equation \(\begin{cases} x + 2 & \text{if } x < 1 \\ x^2 & \text{if } x \geq 1 \end{cases}\).

First, we graph the line \(y = x+2\) until the point \(x = 1\). Then we graph the parabola \(y = x^2\) from \(x = 1\) onward. The important part is to check the boundary at \(x = 1\). If the \(y\)-value of both functions are the same at \(x=1\), then the function is continuous. Otherwise, the function is not continuous and we say there is a discontinuity at \(x = 1\). The graph is shown below: