Course review

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Course review by Mind Map: Course review

1. Working with Quadratic Functions.

1.1. Relating standard form and factored form

1.1.1. Vertex- The point were the parabola changes Max/Min- the lowest or highest point of a parabola X/Y intercept- The point where the parabola goes through the X/Y axis. The standard form of a quadratic equation is y = ax^2 + bx + c, where a, b, and c are coefficients and y and x are variables. It is easier to solve a quadratic equation when it is in standard form because you compute the solution with a, b, and c. However, if you need to graph a quadratic function, or parabola, the process is streamlined when the equation is in vertex form. The vertex form of a quadratic equation is y = m(x-h)^2 + k with m representing the slope of the line and h and k as any point on the line

1.2. Solving problems involving quadratic functions

1.2.1. Let’s say A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H(t)=16t^2+80t+40. Since the word problem is asking for the maximum height of the ball which would be 2.5 seconds., You would try and find the y coordinate of the vertex, The ball would reach the maximum height at the vertex of the parabola. Which is 140m. To find when the ball hits the ground, we need to determine the x intercepts. 5.45. You would use formulas to find the characteristics of the parabola.

2. Introduction to the Quadratic Function

2.1. Characteristics of a function.

2.1.1. On a graph If we can draw any vertical line that intersects a graph more than once, then the graph is not a function because that x value has more than one output. A function has only one output value for each input value. On a mapping diagram if the number on the x axis points to two numbers then it is not a function. And on a table of values or a scatter plot. If the a number of the x axis is repeated twice. Then it is not function. Anything that isn't this is a functiond as the value of f at x or as f of x. Charastericts of a function.

2.1.2. are typically in the form y = mx + b and are graphed as straight lines. To draw a linear graph, start with the y-intercept or b value, then use the slope to find a second point. Quadratic functions are typically in the form y = ax2 + bx + c and are graphed as curved parabolas. Comparing linear and quadratic functions

2.1.3. Consider a linear function y = 3x + 7. To write such function in function notation, we simply replace the variable y with the phrase f(x) to get; f(x) = 3x + 7. This function f(x) = 3x + 7 is read as the value of f at x or as f of x. Working with function notation

2.1.4. if a positive constant is added to a function, f(x)+k, the graph will shift up, Negative moves downward. If a positive constant is added to the value in the domain before the function is applied, f(x+h), the graph will shift to the left. Negative shifts to the right. if the y values are doubled; then the it stretches. If it is divided it’ll shrink flipping upside down, is found by taking the negative of the original function; that is, the rule for this transformation is –f (x). Graphing the quadratic function with transformations

2.1.5. For every polynomial function (such as quadratic functions for example), the domain is all real numbers. if the parabola is opening upwards, i.e. a > 0 , the range is y ≥ k ; if the parabola is opening downwards, i.e. a < 0 , the range is y ≤ k in brackets of { }. The same thing goes for x. If the parabola goes out of the graph and it isn’t possible to decide the value of it then you would write XER or YER. Domain and Range of a quadratic functions

3. Trigonometry and Acute Triangles

3.1. Apply trigonometric ratios

3.1.1. When solving for a side length choose either sin, cos, or tan by determining which side you know and which side you are looking for. ,Substitute your information into the trig ratio, Solve the resulting equation to find the length of the side

3.2. Solve problems involving right triangles

3.2.1. First use the. Pythagorean theorem to find the hypotenuse h of the right triangle. Say we have h = √(6 2 + 8 2) = √(36 + 64) = 10. In a right triangle, using trigonometric ratio for sin(x) we write. sin(x) = 8 / 10 = 0.8. and using trigonometric ratio for cos(x) we write. cos(x) = 6 / 10 = 0.6.

3.3. Solve problems using the sine law

3.3.1. The Sine Rule can be used in any triangle (not just right-angled triangles) where a side and its opposite angle are known. If you need to find the length of a side, you need to use the version of the Sine Rule where the lengths are on the top:You will only ever need two parts of the Sine Rule formula, not all three. You will need to know at least one pair of a side with its opposite angle to use the Sine Rule.

3.4. Solve problems using the cosine law

3.4.1. If you need to find the size of an angle, you need to use the version of the Cosine Rule where the cos(A) is on the left:It is very important to get the terms on the top in the correct order; b and c are either side of angle A which you are trying to find and these can be either way around, but side a must be the side opposite angle A.

3.5. Periodic Behaviour vs sinusoidal

3.5.1. The sinusoidal function family refers to either sine or cosine waves since they are the same except for a horizontal shift. This function family is also called the periodic function family because the function repeats after a given period of time If a function has a repeating pattern like sine or cosine, it is called a periodic function. The period is the length of the smallest interval that

4. Working with Quadratic Functions: Standard form and Factored form

4.1. Converting from vertex form to standard form

4.1.1. When converting to standard form you would want to follow this formula y = a(x - h)^2 + k y = a(x^2 - 2xh + h^2) + k y = ax^2 - 2ahx + ah^2 + k

4.2. Converting from standard form to vertex form by completing the square

4.2.1. Let’s get an example of x^2 + 24x -1 = f(x) Step 1. Group the x's together X^2 + 24 Step 2.Create a space between the second and third term. X^2 + 24 -1 =f(x) Step 3.Now complete the square, by taking the b term (second term) and half it and square it. 24/2=12 and 12^2 =144 Step 4.Take the term from step 3, and write this term, and the negative value of the term in the space created. X^2 +24x +144 -144 -1 =f(x) Step 5.Now group the first three terms together to make our perfect square. (X+12)^2 The perfect square is always the square root of what we just found in step 3. In this example it is square root of 144 which equals 12.Step 6. Add the negative term from step 4 to the constant. (x+12)^2 -145 = f(x)

4.3. -Determining the number of roots using the discriminant

4.3.1. That makes it look like we can always find solutions. However, we can't always take a square root! Sometimes you'll get a negative number under the radical. A negative under the radical means there are no real number solutions to the radical. We can use the formula under the radical, B^2 −4ac, also known as the discriminant to determine the number of roots of solutions in a quadratic equation. We can use the formula under the radical, b2−4ac, also known as the discriminant to determine the number of roots of solutions in a quadratic equation. There are three cases: b^2−4ac<0: The equation has 0 real solutions b^2−4ac=0: The equation has 1 real solution. b^2−4ac>0: The equation has 2 real solutions.

4.4. -Solving quadratic equations using the Quadratic formula

4.4.1. in order to solve a quadratic equation by the Quadratic Formula, we first need to have the quadratic equation in standard form, i.e., ax² + bx + c = 0, where a, b, and c are real numbers and a does not equal 0. The Quadratic Formula is: x = [‒b ± √(b² ‒ 4ac)]/(2a). The a, b, and c in the Quadratic Formula match or correspond to the a, b, and c coefficients in the quadratic equation in standard form: ax² + bx + c = 0, and that's because the Quadratic Formula is derived from the quadratic equation. So, once you get the quadratic equation to be solved into standard form, it's just a simple matter of identifying the numerical values for a, b, and c from the equation and then substituting those values into the Quadratic Formula for the corresponding a, b, and c and then simplifying and solving for x. Example. x = [‒b ± √(b² ‒ 4ac)]/(2a) = [‒5 ± √(5² ‒ 4(1)(6))]/[2(1)] = [‒5 ± √(25 ‒ 24)]/2 = [‒5 ± √1]/2 = [‒5 ± 1]/2 Therefore, our solutions for the given quadratic equation are: x= [‒5 + 1]/2 = ‒4/2 = ‒2 and x = [‒5 ‒ 1]/2 = ‒6/2 = ‒3

5. Algebra of Quadratic Expressions

5.1. Multiply two binomials

5.1.1. When multiplying binomials you. Multiply in order like this, the First terms. Multiply the Outer terms. Multiply the Inner terms. Multiply the Last terms. Combine like terms, when it is possible.

5.2. Expanding and simplifying expressions

5.2.1. To expand a bracket means to multiply each term in the bracket by the expression outside the bracket. For example, in the expression 3 ( m + 7 ) , multiply both and 7 by 3, so: 3 ( m + 7 ) = 3 × m + 3 × 7 = 3m + 21. And for simplifying. Remove any grouping symbol such as brackets and parentheses by multiplying factors. use the exponent rule to remove grouping if the terms are containing exponents. Combine the like terms by addition or subtraction. Combine the constants

5.3. Factoring simple trinomials

5.3.1. factor a trinomial in the form x2 + bx + c, find two integers, r and s, whose product is c and whose sum is b. Rewrite the trinomial as x2 +rx + sx + c and then use grouping and the distributive property to factor the polynomial. The resulting factors will be (x + r) and (x +

5.4. Factoring complex trinomials

5.4.1. Remove Common Factors if possible.If the coefficient of the x2 term is 1, then. x2 + bx + c = (x + n)(x + m), where n and m. · Multiply to give c. Add to give b. If the coefficient of the x2 term is not 1, then use either. Guess-and Check. List the factors of the coefficient of the x2 term. The you’d factor by grouping.

5.5. Factoring special cases (perfect square and difference of squares)

5.5.1. If you’re going to factor a perfect square first you would do is Factor out the GCF, if necessary. Write each term as a perfect cube. Identify the given variables and the terms of the binomial are the cube roots of the terms of the original polynomial. Every difference of squares problem can be factored as follows: a2 – b2 = (a + b)(a – b) or (a – b)(a + b). So, all you need to do to factor these types of problems is to determine what numbers squares will produce the desired results. After that you’d determine if the remaining factors can be factored any further