# FTCE Mathematics 6-12: Ultimate Guide and Practice Test

Preparing to take the FTCE Mathematics 6-12 exam?

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### FTCE Mathematics 6-12 Quick Facts

Florida teacher certificate candidates are required to pass this test which covers mathematics for grades 6-12.

There are 75 multiple-choice questions, broken up into graphic, direct question, sentence completion, command, word problem, and selection items.

**Cost: **

$150

**Scoring: **

A scaled score of at least 200 is needed to pass the exam.

**Pass rate: **

The pass rate is 67%.

**Study time: **

In order to pass, you should break the test topics down into your strengths and weaknesses. Give yourself enough time to work on each topic until you feel confident with the majority of the subjects. Create a study plan detailing the time you need to study each topic and the days you plan to do that.

**What test takers wish they would’ve known: **

- A Texas Instruments TI-30X IIS scientific calculator is provided at the test site. Examinees may not bring their own calculator.
- Unofficial pass/non-pass status is provided immediately after testing.
- You will not be penalized for incorrect responses, so make an educated guess if you are not sure.

Information and screenshots obtained from the National Evaluation Series website: http://www.fl.nesinc.com/prepPage.asp?test=026

## Exam Content

### Overview

The exam has ten competencies:

- Algebra (13%)
- Advanced Algebra (12%)
- Functions (8%)
- Geometry (15%)
- Coordinate Geometry (6%)
- Trigonometry (7%)
- Statistics and Probability (10%)
- Calculus (9%)
- Mathematical Reasoning (5%)
- Instruction and Assessment (15%)

So, let’s talk about Algebra first.

### Algebra

This competency includes about 10 multiple-choice questions which makes up about 13% of the entire exam.

These questions test your knowledge of linear, absolute value, rational, radical, and quadratic equations, as well as the properties of real numbers.

Let’s talk about some concepts that you will more than likely see on the test.

**Solving Inequalities Involving Absolute Value**

Example 1: Solve the inequality |3x-6>15|.

We can rewrite this inequality as two inequalities without the absolute value as follows:

3*x* – 6 < -15 or 3*x* – 6 > 15

Now, solve both of these inequalities separately by isolating *x*:

3x < -15 + 6 | or | 3x > 15 + 6 |

3x < -9 | or | 3x > 21 |

x < -3 | or | x > 7 |

Now, we can graph this solution on a number line:

Example 2: Solve the inequality |2x+5< 11|.

We can rewrite this inequality as two inequalities without the absolute value as follows:

-11 < 2x + 5 < 11 |

-11 – 5 < 2x < 11 – 5 |

-16 < 2x < 6 |

-8 < x < 3 |

Now, we can graph this solution on a number line:

**Completing the Square**

To solve a quadratic equation by completing the square, the goal is to be able to factor one side of the equation as a perfect square binomial, and then solve using the square root method. To do this, we follow these steps:

- Put the equation x
*²*+bx+c=0 in the form*x²**+bx = -c*. - Add the term (b/2)
*²*to both sides of the equation. - Factor the left-hand side as (x+b/2)
*²*. - Take the square root of both sides and solve for
*x*.

For example, consider the quadratic equation x*²*+6x – 3 = 0.

Step 1: Add 3 to both sides to put the equation into the correct form:

x*²*+6x = 3

Step 2: Add the term (b/2)*² *= (6/2)*² *= 9to both sides of the equation:

x*² *+ 6x + 9 = 3+9

x*² *+ 6x + 9 = 12

Step 3: Factor the left-hand side:

(x + 3)*² *= 12

Step 4: Solve for *x*:

x+3* *= √12

x= -3 ± √12

These are the solutions of the equation.

### Advanced Algebra

This competency includes about 9 multiple-choice questions which make up about 12% of the entire exam.

These questions test your knowledge of polynomials, exponents, logarithms, vectors, sequences, matrices, and complex numbers.

Here are some concepts you should know.

**Polynomial Equations**

To solve higher degree polynomials (degree 3 or higher), we can apply the rational root theorem which says that if pqis a zero of a polynomial, then *p* is a factor of the constant term of the polynomial, and *q* is a factor of the leading coefficient of the polynomial.

Now we can use synthetic division to determine which of these are in fact roots.

We still can’t factor this cubic polynomial, so we need to use synthetic division again, but this time, let’s try with our new polynomial. Let’s try *x* = 1:

**Matrices**

Matrices are rectangular arrays of numbers placed in rows and columns inside brackets. They are used to solve systems of linear equations quickly.

There are three types of row operations you can perform on a matrix:

- Two rows may be interchanged.
- Any row can be multiplied by a non-zero number.
- A multiple of a row can be added to another row.

To solve a system of linear equations using these row operations, begin by writing the augmented matrix for the system by taking the equations in standard form and placing the coefficients in the equations to the left of a vertical line and the constants to the right. Then use row operations until the matrix to the left of the vertical line consists of 1s down the diagonal from upper left to lower right and 0s below the 1s.

Then a new system of equations can be written from this new augmented matrix, which gives the solutions to the system.

For example, let’s solve the system:

*x* + *y* – *z* = -2

2*x* – *y* + *z* = 5

–*x* + 2*y* + 2*z* = 1

We can write the augmented matrix:

Replace Row 3 with Row 1 plus Row 3:

Replace Row 3 with Row 1 plus Row 3:

Replace Row 2 with -2 times Row 1 plus Row 2:

Multiply Row 2 by –1/3:

Replace Row 3 by -3 times Row 2 plus Row 3:

Multiply Row 3 by 1/4:

Replace Row 2 with Row 3 plus Row 2:

Replace Row 1 with Row 3 plus Row 1:

Now, the 3 by 3 matrix on the left has a diagonal of 1s with 0s below it. This means we can stop performing row operations. Writing a new set of equations from this matrix gives:

*x – z = -1*

* y = -1*

*z = 2*

Substituting *z* = 2 into the first equation gives *x* – 2 = -1. Therefore, *x* = 1.

The solution to the system is *x* = 1, *y* = -1, and *z* = 2.

### Functions

This competency includes about 6 multiple-choice questions which make up about 8% of the entire exam.

These questions test your knowledge of functions including their domain and range, graphs, inverses, and transformations.

The following concepts may pop up on the test.

**Finding the Inverse of a Function**

Suppose you have the function *f*(*x*) = √(3x-1). How do you find the inverse function f¯¹(x)?

Step 1: Start by substituting the variable *y* for *f*(*x*):

*y* = √(3x – 1)

Step 2: Switch the variables *x* and *y* in the equation:

*x* = √(3y – 1)

Step 3: Solve for the variable *y*:

*x² **=√(3y – 1)*

*x² **+ 1 = 3y*

*(x² **+ 1/**3) **= y*

Therefore, f¯¹(x) = (*x²**+1/**3)*.

**Symmetric, Periodic, and Even/Odd Functions**

*Symmetry*

A function can have three kinds of symmetry:

*x*-axis symmetry: –*f*(*x*) = *f*(*x*) for all *x. *

*y*-axis symmetry: *f*(*x*) = *f*(-*x*) for all *x. *If there is *y*-axis symmetry, then we call the function even.

Origin symmetry: –*f*(*x*) = *f*(-*x*) for all *x.* If there is *x*-axis symmetry, then we call the function odd.

Example 1: Consider the function *f*(*x*) = 5*x² **+ 1*. Let’s test for symmetry:

*f*(-*x*) = 5*(-x)² **+ 1 = *5x² + 1 = *f*(*x*)

Therefore, this function has *y*-axis symmetry and is even.

–*f*(*x*) = *–*5x*² *– 1 ≠ 5x*² *+ 1 = *f*(-*x*)

Therefore, this function is not symmetric over origin.

–*f*(*x*) = *–*5x*² *– 1 ≠ 5x*²* + 1 = *f*(*x*)

Therefore, this function is not symmetric over the x-axis.

*Periodicity*

A function is periodic if there is some real number T such that *f*(*x*) = *f*(*x* + T) for all real numbers *x*. In other words, the function repeats itself. This is often easier to see graphically.

The most well known periodic functions are trigonometric functions such as sin *x*, cos *x*, and tan *x*; however, all expressions containing these trig functions are not periodic.

For example, *f*(*x*) = sin(*x²**)*is graphed below and does not repeat itself, so it is not periodic.

However, g(*x*) = 3sin(x) + 6 repeats with a period of 2, since

*g*(*x*) = 3sin(x) + 6 = 3sin(x + 2π) + 6 = *g*(x + 2π) for all *x*.

### Geometry

This competency includes about 11 multiple-choice questions which makes up about 15% of the entire exam.

These questions test your knowledge of circles, triangles, quadrilaterals, and other polygons, surface area, and volume, as well as geometric constructions.

Let’s talk about some concepts that you will more than likely see on the test.

**Properties of Quadrilaterals**

**Finding the Measure of Angles in Convex Polygons**

Theorem: If a convex polygon has *n* sides, then the sum of the interior angles of the polygon is (*n* – 2)*180°.

Theorem: If a convex polygon has *n* sides, the sum of the exterior angles of the polygon is 360°.

For example, consider the diagram below. Solve for *x* and *y*.

This diagram has a pentagon and a triangle contained in a larger 7-gon. The interior angle of the pentagon has a sum of (*n* – 2)*180° = (5 – 2)*180° = 540°.

Using the fact that the exterior angle of 45° is supplementary to one of the interior angles, we can find that the missing interior angle of the pentagon is 135°.

Then:

56 + 124 + 63 + (*y* – 12) + 135 = 540

378 + (*y* – 12) = 540

*y *= 174

Now, *x* + (174 – 12) = 180, since the angles are supplementary; therefore, *x* = 18.

### Coordinate Geometry

This competency includes about 5 multiple-choice questions which makes up about 6% of the entire exam.

These questions test your knowledge of the coordinate plane including conic sections, transformations of figures, and distance and midpoint formulas.

Here are some concepts you need to know.

**Determining the Equation of Parabolas**

To find the equation of a parabola, we need to know two things. First, we need the vertex of the parabola. Second, we need another point on the parabola.

For example, consider the parabola below:

Use the vertex form of a parabola *y=a**x-h**2**+k*, where (*h*, *k*) is the vertex of the parabola.

In this example, the vertex is the point (4, 3). Therefore, *h* = 4 and *k* = 3. The equation is *y=a**x-4**2**+3*.

Now, choose another point on the equation. We choose the point (3, 1). Substitute the point into the equation for *x* and *y*:

*1 = a (**3 – 4)²** **+ 3*

*1 = a(**-1)**² **+ 3*

*1 = a + 3*

-2 = a

Therefore, the parabola is *y= -2(**x-4)**² **+ 3*.

**Determining the Equation of a Conic Section**

There are several different kinds of conic sections. Parabolas, which we have already discussed above, are formed by the intersection of a right circular cone and a plane.

Ellipses are represented by one of the two equations below depending on if its major axis is parallel to the *x*-axis or the *y*-axis.

Also, the vertices occur at the points (*h* – *a*, *k*) and (*h* + *a*, *k*) when the major axis is parallel to the *x*-axis and (*h*, *k* – *a*) and (*h*, *k* + *a*) when the major axis is parallel to the *y*-axis.

For example, consider the ellipse below:

This ellipse has a center point at (*h*, *k*) = (-1, 2). We can use the second form of the equation, since the major axis is parallel to the *y*-axis. The vertices of the parabola occur at the points (-1, -2) and (-1, 5). Therefore, (*h*, *k* – *a*) = (-1, -2) which means that *k* – *a* = 2 – *a* = -2. Solving for *a* gives *a* = 4.

Therefore, the equation is written as:

Now, plug in any other point on the ellipse to solve for the value of *b*. We choose (2, 2):

Hyperbolas are another type of conic section. To find the equation of a hyperbola, you need the equations of its asymptotes, as well as vertices.

### Trigonometry

This competency includes about 5 multiple-choice questions which makes up about 7% of the entire exam.

These questions test your knowledge of trigonometric functions including their graphs, applications to the real-world, and identities.

Check out this important concept.

**Trigonometric Ratios**

There are 6 trigonometric ratios that you should know:

When given x, an angle measure in a right triangle, and two of the sides opposite and adjacent to the angle or the hypotenuse, this provides the information needed to find the length of the missing side.

Example 1: Find the hypotenuse of the triangle below.

First, consider which trigonometric identity is appropriate to use. Here we are given an angle and the opposite side, and we would like to know the hypotenuse. The appropriate trig identity is the sine function, since it involves the opposite side and the hypotenuse.

### Statistics and Probability

This competency includes about 8 multiple-choice questions which makes up about 10% of the entire exam.

These questions test your knowledge of probability and statistics, including counting principles, bias, expected value, charts and graphs, linear regression, and statistical measures.

Here are some concepts you definitely need to know.

**Measures of Central Tendency**

Measures of central tendency help us to determine how data is distributed by identifying a single value at roughly the center of the data. We will consider several measures of central tendency below.

The mean of a data set is the average value of a data set. This can be found by adding together all of the values and dividing by the total number of values.

The mode of a data set is the value that occurs most frequently in the data set. It is possible to have more than one mode in a data set if several values occur the most.

The median of a data set is the middle value in the set. If there is an even number of data points, there will not be an exact middle. In this case, the median is found by taking the average of the two data points closest to the middle.

For example, suppose that the ages for a group of ten students were collected and are listed below:

9, 11, 13, 11, 8, 7, 13, 9, 9, 12

The mean of this data set can be found by adding all of these ages together and dividing by 10, since there are 10 students:

Mean = (9+11+13+11+8+7+13+9+9+12) / 10 = 10.2

To find the mode and median of a data set, it is helpful to reorder the set from lowest to highest:

7, 8, 9, 9, 9, 11, 11, 12, 13, 13

Now, we can see that the mode of the data set is 9, since 9 occurs 3 times, which is more than any other data point.

Since the data set has an even number of values, there are two values in the middle: 9 and 11. To find the median, you must average 9 and 11; therefore, the median of the data set is 10.

**Permutations and Combinations**

A permutation is an arrangement of items that occurs when the order of the items makes a difference, and no item is used more than once.

Usually, we are concerned with how many permutations can be made from a collection of items.

The formula for the number of permutations of *n* things taken *r* at a time is denoted as nPr:

nPr=n!(n-r)!

For example, suppose that you have 20 different people in your family, and you want to know how many ways there are to fill 5 seats on a couch, where the order of the people on the couch matter.

Then:

So, there are 1,860,480 different ways to fill the 5 seats.

A combination is a collection of items in which the order does matter, but no item is used more than once.

The formula for the number of combinations of *n* things taken *r* at a time is denoted as nCr:

For example, suppose you have a 52 card deck, and you deal 7 cards. How many different hands of 7 cards are possible? Since it does not matter what order the cards are in for your hand, we need to count the combinations of 52 things taken 7 at a time:

So, there are 133,784,560 possible combinations of 7 cards chosen from the deck.

### Calculus

### Mathematical Reasoning

This competency includes about 4 multiple-choice questions which makes up about 5% of the entire exam.

These questions test your knowledge of mathematical reasoning including counterexamples, analyzing proofs, and inductive versus deductive reasoning.

Take a look at this important concept.

**Inductive versus Deductive Reasoning**

Deductive reasoning, sometimes called top-down reasoning, starts with premises that are known to be true and continues towards a logical conclusion.

For example, humans breathe air. David is a human; therefore, David breathes air.

Inductive reasoning, sometimes called bottom-up reasoning, starts with specific observations and combines them to make broad generalizations.

For example, you notice that Julia shows up late to work. Julia is a teenager; therefore, you conclude that teenagers are irresponsible.

Conclusions reached using inductive reasoning can be wrong sometimes, but this type of reasoning is still important. Inductive reasoning is most often used to form hypotheses, which can then be tested using the scientific method.

### Instruction and Assessment

This competency includes about 11 multiple-choice questions which makes up about 15% of the entire exam.

These questions test your knowledge of sound classroom theory and practices including lesson sequencing, mathematical models, problem-solving strategies, learning environments, and assessments.

The following concepts may pop up on the test.

**Types of Assessments**

Formative assessment is given regularly throughout the school year to provide ongoing feedback and help teachers determine concepts that need to be covered in more detail. Formative assessments also help students identify where their own understanding can be improved. Formative assessment is usually lower stakes and can include very informal tasks such as games, projects, and group work, as well as writing a summary about the main point of the lesson that day or submitting an outline of a paper before writing it.

Diagnostic assessment is given before the school year starts, or before a unit or activity, to determine the level of prior knowledge and understanding a student already has of a subject. Often these types of assessments aren’t given a grade but are simply used to inform the teacher. This might look like a short quiz or a warm-up activity.

Summative assessments are usually given at the end of a chapter, unit, or course to determine how much the student has learned and retained. These types of assessments are usually higher stakes and include tests, quizzes, final papers, and cumulative projects.

**Problem-Solving Strategies**

Some problem-solving strategies for mathematics include working backward, looking for a pattern, making an estimate, guess and check, and drawing a picture.

For example, suppose Penny bought some cookies at a bake sale. She ate 3 of them and gave 2 of them to her sister. After dividing the rest equally with her brother, she had 8 cookies remaining. How many cookies did she start with?

This problem can be solved by working backward from the final number of cookies she has at the end of this process. She ends with 8 cookies, which was half of what she had before she shared with her brother. At that point, she had 16 cookies. She gave 2 cookies to her sister. So, she had 18 cookies before doing that. She ate 3 cookies first; therefore, she started with 21 cookies.

Guess and check works well with finding the roots of polynomial functions. In order to determine if a value is a root of the function, you must make a reasonable guess, and then apply synthetic division to determine if the remainder of the division is 0. If so, then you have found a root.

And that’s some basic info about the FTCE Mathematics 6-12 exam.