How This Exam Is Built
The B.E.S.T. Algebra 1 End-of-Course (EOC) Assessment is a computer-adaptive test taken in one 160-minute session, with a short break after the first 80 minutes. It has 45 to 50 questions, and in the spring about 5 of them are field-test questions that do not count. Because the test is adaptive, the questions you see depend on how you answer the earlier ones.
An online scientific calculator and an online reference sheet are available for the whole test, and you may also be allowed to use an approved handheld scientific calculator (graphing calculators and calculators that solve equations are not allowed). The reference sheet lists the three forms of a linear equation, the three forms of a quadratic function, the forms of an exponential function, the quadratic formula, the simple and compound interest formulas, and unit conversions, so learn how to use them rather than memorizing them. Questions come in several formats, including multiple choice, multiselect (select all that apply), equation editor (you type the answer), matching tables, drop-down choices, and graphing. Some items have more than one part or more than one answer box. Our B.E.S.T. Practice Exam uses multiple choice, multiselect, typed answers, and matching; it does not include drag-and-drop graphing, so practice plotting points and lines on paper too.
Scores run from 325 to 475. Level 3, which starts at 400, means on grade level, and 400 is the passing score for graduation. If you take the EOC as part of your Algebra 1 course, it also counts for 30 percent of your course grade. The official blueprint splits the test into three reporting categories of about equal weight, each 31 to 38 percent: Expressions, Functions, and Data Analysis; Linear Relationships; and Non-Linear Relationships. The sections below follow those categories.
Expressions, Functions, and Data Analysis
Exponents and Radicals
A rational exponent combines a root and a power: \(x^{\frac{m}{n}}\) \(= \left(\sqrt[n]{x}\right)^{m}\). Use the exponent rules to simplify: add exponents when multiplying the same base, subtract when dividing, and multiply when raising a power to a power. A negative exponent means a reciprocal. To add or subtract radicals, simplify each one first and combine only like radicals.
Worked example. Simplify \(\frac{\left(2x^{3}\right)^{2}}{4x^{-1}}\). The numerator is \(4x^{6}\). Dividing, \(\frac{4}{4} = 1\) and \(x^{6 - (-1)} = x^{7}\), so the result is \(x^{7}\).
Common mistake. Adding the numbers under radicals, as in \(\sqrt{18} + \sqrt{8}\) \(= \sqrt{26}\). Simplify instead: \(3\sqrt{2} + 2\sqrt{2}\) \(= 5\sqrt{2}\).
Practice next. B.E.S.T. Practice Exam, Questions 1, 13, and 31.
Reading Expressions and Rearranging Formulas
Many questions describe a situation and ask what part of an expression means. Read each part for what it counts or measures and what units it has. To solve a formula for a variable, undo the operations around that variable in reverse order.
Worked example. Solve \(V = \frac{1}{3}\pi r^{2}h\) for \(h\). Multiply both sides by 3: \(3V = \pi r^{2}h\). Divide by \(\pi r^{2}\): \(h = \frac{3V}{\pi r^{2}}\).
Common mistake. Undoing operations in the wrong order. When a sum is multiplied, as in \(2A\) \(= h\left(b_{1} + b_{2}\right)\), divide by \(h\) before you subtract \(b_{1}\). When a number is added to a product, as in \(y = mx + b\), subtract \(b\) before you divide by \(m\).
Practice next. B.E.S.T. Practice Exam, Questions 4 and 22.
Functions and Their Key Features
Function notation \(f(a)\) means the output when the input is \(a\). The average rate of change from \(x = a\) to \(x = b\) is \(\frac{f(b) - f(a)}{b - a}\). From a table with equally spaced inputs, constant first differences mean a linear function, constant second differences mean a quadratic, and a constant ratio means an exponential function. Adding \(k\) outside a function, \(f(x) + k\), shifts the graph up or down, while \(f(x + k)\) shifts it left or right, opposite to the sign of \(k\).
Worked example. For \(g(x) = 3x^{2} - x\), find the average rate of change from \(x = 1\) to \(x = 3\). \(g(1) = 2\) and \(g(3) = 24\), so the rate is \(\frac{24 - 2}{3 - 1} = 11\).
Common mistake. Calling a function exponential just because its outputs grow faster and faster. A quadratic does that too; check for a constant ratio.
Practice next. B.E.S.T. Practice Exam, Questions 7, 16, 19, 28, 34, and 43.
Data: Distributions, Two-Way Tables, and Samples
Compare box plots by median (center) and interquartile range (spread). In a two-way table, a conditional relative frequency divides by the total of the given condition. A sample percent with a margin of error gives an interval of percents; multiply both ends by the population size to estimate a total.
Worked example. In a sample, 45% of voters support a plan, with a margin of error of 3 percentage points. For a town of 8,000 voters, the estimate is \(0.42 \cdot 8{,}000 = 3{,}360\) to \(0.48 \cdot 8{,}000 = 3{,}840\) supporters.
Common mistake. Using the wrong total in a two-way table. Read the question for the words that name the group, such as "of the students in the band," and divide by that group's total.
Practice next. B.E.S.T. Practice Exam, Questions 10, 25, and 37.
Simple Versus Compound Interest
Simple interest adds the same amount every year, so the balance grows linearly: \(A = P(1 + rt)\). Compound interest is earned on the growing balance, so the balance grows exponentially: \(A\) \(= P\left(1 + \frac{r}{n}\right)^{nt}\).
Worked example. $500 at 6% simple interest earns $30 every year. At 6% compounded annually it is multiplied by 1.06 each year, so it earns $30.00, then $31.80, then about $33.71.
Common mistake. Treating both kinds of interest as adding the same amount each year. Only simple interest does that.
Practice next. B.E.S.T. Practice Exam, Question 40.
Linear Relationships
Writing and Solving Linear Equations and Inequalities
Write an equation by naming the variable and translating each quantity: a rate multiplies the variable, and a one-time amount is added. Solve by undoing operations. For inequalities, the steps are the same except that multiplying or dividing by a negative number reverses the sign. In a compound inequality, apply each step to all three parts.
Worked example. Solve \(4 - 3x \ge -11\). Subtract 4: \(-3x \ge -15\). Divide by \(-3\) and flip the sign: \(x \le 5\).
Common mistake. Forgetting to reverse the sign after dividing by a negative. Test a value to check: \(x = 0\) gives \(4 \ge -11\), which is true, and 0 is in \(x \le 5\).
Practice next. B.E.S.T. Practice Exam, Questions 2 and 11.
Lines: Slope, Forms, and Key Features
Slope is \(\frac{\text{change in } y}{\text{change in } x}\). The reference sheet gives slope-intercept, standard, and point-slope forms. Parallel lines have equal slopes, and perpendicular slopes are opposite reciprocals. In a model, the slope is the rate of change and the \(y\)-intercept is the starting value; the situation decides the reasonable domain and range.
Worked example. Write the line through \((2, -1)\) perpendicular to \(y = 4x - 3\). The perpendicular slope is \(-\frac{1}{4}\), so \(y + 1 = -\frac{1}{4}(x - 2)\), or \(y\) \(= -\frac{1}{4}x - \frac{1}{2}\).
Common mistake. Using the reciprocal without changing the sign, which gives a slope of \(\frac{1}{4}\). Perpendicular slopes multiply to \(-1\).
Practice next. B.E.S.T. Practice Exam, Questions 5, 8, 26, 38, and 44.
Two-Variable Inequalities and Systems
To graph a two-variable inequality, draw the boundary line (dashed for \(<\) or \(>\), solid for \(\le\) or \(\ge\)) and shade the side that makes the inequality true. A system of equations is solved by graphing, substitution, or elimination. The solution set of a system of inequalities is the overlap of the shaded regions. Real-world limits become constraints such as \(20x + 30y \le 600\) and \(x \ge 0\).
Worked example. Solve \(x + y = 30\) and \(5x + 2y = 96\). Substitute \(y = 30 - x\): \(5x + 60 - 2x = 96\), so \(3x = 36\), \(x = 12\), and \(y = 18\).
Common mistake. Counting a point on a dashed boundary as a solution. If the boundary is dashed, its points make the inequality false.
Practice next. B.E.S.T. Practice Exam, Questions 14, 17, 23, 32, and 35.
Scatter Plots, Lines of Fit, and Residuals
A line of fit follows the trend of a scatter plot, with about as many points above as below. Its slope gives the direction of the correlation, and the residuals (actual minus predicted values) show the strength: small residuals mean a strong correlation. A correlation alone never proves that one quantity causes the other.
Worked example. A line of fit is \(y = 1.8x + 12\). For \(x = 10\) it predicts \(y = 30\). If the actual value is 33, the residual is \(33 - 30 = 3\).
Common mistake. Reading a strong correlation as proof of cause and effect. Look for a third factor that could drive both quantities.
Practice next. B.E.S.T. Practice Exam, Questions 20, 29, and 41.
Non-Linear Relationships
Polynomials and Factoring
To multiply polynomials, multiply every term by every term and combine like terms. To divide by a monomial, divide each term. To factor completely, take out the greatest common factor, then look for grouping, a difference of squares, or a factorable trinomial, and keep going until nothing factors further.
Worked example. Factor \(3x^{2} - 48\) completely. The GCF is 3: \(3\left(x^{2} - 16\right)\). Then \(x^{2} - 16\) is a difference of squares, so the answer is \(3(x - 4)(x + 4)\).
Common mistake. Stopping after the GCF or after grouping. Always check whether a factor such as \(x^{2} - 16\) can be factored again.
Practice next. B.E.S.T. Practice Exam, Questions 3, 15, and 30.
Quadratic Equations and Functions
Solve quadratic equations by factoring, square roots, completing the square, or the quadratic formula. The three forms of a quadratic each show something: standard form shows the \(y\)-intercept, vertex form shows the vertex, and factored form shows the zeros. The vertex is at \(x = -\frac{b}{2a}\) and is halfway between the zeros. In context, the vertex gives a maximum or minimum, and the zeros give where a quantity equals 0. When a model gives two solutions, keep only the one that makes sense in the situation, such as a positive time.
Worked example. A ball's height is \(h(t) = -16t^{2} + 32t + 6\). The vertex is at \(t = \frac{-32}{-32} = 1\), so the maximum height is \(h(1) = -16 + 32 + 6 = 22\) feet.
Common mistake. Answering with the time when a question asks for the height, or the reverse. Reread what the question asks for before you enter the answer.
Practice next. B.E.S.T. Practice Exam, Questions 6, 12, 21, 27, 36, and 42.
Absolute Value
Solve \(\lvert \text{expression}\rvert\) \(= c\) by isolating the absolute value and splitting into two equations. The graph of \(f(x) = a\lvert x - h\rvert + k\) is a V with vertex \((h, k)\); it opens up if \(a > 0\) and down if \(a < 0\).
Worked example. Solve \(2\lvert x + 3\rvert - 4 = 10\). Then \(\lvert x + 3\rvert = 7\), so \(x + 3 = 7\) or \(x + 3 = -7\), and \(x = 4\) or \(x = -10\).
Common mistake. Splitting into two cases before isolating the absolute value. Undo the \(-4\) and the 2 first.
Practice next. B.E.S.T. Practice Exam, Questions 18 and 39.
Exponential Functions
An exponential function multiplies by the same factor for every unit increase in \(x\). In \(f(x) = ab^{x}\), \(a\) is the starting value and \(b\) is the factor: growth if \(b > 1\), decay if \(0 < b < 1\). In \(a(1 \pm r)^{x}\), \(r\) is the percent rate of change. The graph of \(f(x) = ab^{x}\) has no \(x\)-intercept, and its range is \(y > 0\) when \(a > 0\).
Worked example. A car worth $24,000 loses 15% of its value each year: \(V(t)\) \(= 24{,}000(1 - 0.15)^{t}\) \(= 24{,}000(0.85)^{t}\). After 2 years, \(V(2)\) \(= 24{,}000(0.7225)\) \(= 17{,}340\) dollars.
Common mistake. Writing the decay rate as the base, as in \(24{,}000(0.15)^{t}\). The base is the part that remains, \(1 - 0.15 = 0.85\).
Practice next. B.E.S.T. Practice Exam, Questions 9, 24, 33, and 45.