By Harkeerat Kanwal
Unit 1: Vertex Form
- Use a table of differences to determine whether an equation is linear or quadratic.
- Graph a parabola using an equation in vertex form.
- Apply vertical stretch or compression to a parabola when graphing using the step pattern.
- Determine the equation for a parabola given the vertex and one point.
y = a(x - h)² + k
The vertex of a parabola is the highest or lowest point of the curve and can be determined from the equation using the values of h and k where h is the x value of the vertex and k is the y value. The a value is the vertical stretch or compression of the parabola where an absolute value less than 1 indicates a compression while an absolute value greater than or equal to 1 indicates a stretch.
Table of Differences
If the numbers in the first differences column are not the same, however, create another new column called second differences (coloured purple in table below). Do the same steps as before but this time subtract from the values in the first differences column
(-4 - 14 = 10), (6 - (-4) = 10), and so forth. If the values of the second differences are all the same then the equation is quadratic.
To sum it up if the first differences are the same then the equation is linear, if the second differences are the same then the equation is quadratic and if neither are the same then it is neither linear nor quadratic.
y = 2(x - 4)² - 6
The value of h in this equation is 4 so the x value of the vertex is 4. Keep in mind that it is not -4 because the negative is already present in the original form of the equation. Next we see that the value of k is -6 so the y value of the vertex is -6. Knowing these coordinates we can now plot the vertex at point (4, -6).
Next, we apply the vertical stretch by using the step pattern. In the case where there is no stretch or compression applied, like in the simple equation y = x², the steps are: 1, 3, 5, 7 and so on. This means that from the vertex you go over 1 up 1 unit, then up 3 units, then up 5 units and so on.
In the graph below we have to use a vertical stretch of 2 so we multiply each step in the vanilla step pattern by 2 to get the new steps. This results in the steps: 2, 6, 10, 14, and so on.
Finding the Equation
To find the value of a is not as complicated as you may first suspect. Simply substitute in a point of the parabola by subbing in its x and y values and solve for a.
Q. Determine the equation for a parabola that has a vertex of (2, -4) and passes through the point (-6, -20).
A. To solve these kinds of questions first set up the equation with the vertex points:
y = (x - 2)² - 4
The x coordinate of the vertex is 2 so we sub in 2 for h. The y coordinate for the vertex is -4 so we substitute in -4 for k. Next, we'll sub in the x and y coordinates of the point we know the parabola passes through, which is (-6, -20), in order to solve for a:
-20 = a(-6 - 2)² - 4
-0.25 = a
Solving the equation we find that the value of a is -0.25 so we can plug this into our original equation to get the final answer:
y = -0.25(x - 2)² - 4
Therefore the question for this parabola is y = -0.25(x - 2)² - 4.
Unit 2: Factored Form
- Simplify expressions using distributive property with the FOIL method.
- Factor polynomials using common factors and by grouping.
- Factor simple and complex trinomials using decomposition.
- Factor using difference of squares and perfect square trinomials.
- Solve a quadratic equation in factored form.
- Graph a parabola from the factored form of its equation.
y = a(x - r)(x - s)
The x-intercepts or zeros of a parabola are points that occur when y = 0. The zeros of the parabola defined by this equation can be determined by setting y to 0 and solving for x. Since every parabola has two zeros there are two x in the equation so solving for both of them will give you both zeros. The value of a, like in vertex form, is the vertical stretch or compression that is applied to the parabola.
The term "distributive property" may seem daunting when you first encounter it but, in truth, it's quite simple. It just means that when you are doing the following kind of multiplication, all of the terms in the first pair of brackets must be multiplied by all of the terms in the second pair of brackets:
(x + 5)(x - 1)
To simplify this equation, you can use FOIL to multiply all of the terms in the bracket. FOIL stands for "First, Outside, Inside, and Last". Following this method, to simplify, first multiply the "First" terms in each pair of brackets together. In this case, the two first terms are both x. Multiplying them together, we get x². Next we multiply the two "Outside" terms together, in this case they are x and -1, which gives us -x. Then we multiply the two "Inside" terms, they are 5 and x, giving us 5x. Finally, we multiply the last two terms, 5 and -1, giving us -5.
After using FOIL to multiply all of the terms in the brackets, we get the simplified form:
= x² - x + 5x - 5
= x² + 4x - 5
Here is a picture to visualize the FOIL method in action:
Common Factoring and Grouping
4x²y + 6xy²
In the above example we can see that a 2xy can be factored out from the terms. This is because both the coefficients have a GCF of 2 and both of the variables of the terms have a GCF of xy. The resulting expression is:
= 2xy(2x + 3y)
If you have two binomials that are exactly the same you can think of them as a common factor:
4x(w + 1) + 5y(w + 1)
In this example both terms have a common factor or w + 1. You can think of this as no different than any other variable that could have been there. Following the rules of common factoring that we've covered above, the GCF in this expression can be factored out:
= (w + 1)(4x+ 5y)
Finally there is something called factoring by grouping which is can be used when you have a polynomial with 4 terms. It is simply when you factor something out of each pair of terms and then use the above method to factor out a GCF.
x² + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
Factoring using Decomposition
Trinomials can be in one of two forms: simple or complex. A simple trinomial is when the x² has a coefficient of 1 while a complex trinomial is the the x² has a coefficient not equal to 1. If the video below you will learn how to solve both kinds.
64x² - 81
To solve use the following expression where a is the square root of the coefficient of the first term and b is the square root of the second term:
(ax - b)(ax + b)
So if we substitute in the values of a and b into the expression we get our answer which is:
(8x - 9)(8x + 9)
The second special case, a perfect square trinomial, can be recognized when the coefficient of the first term and the third term are perfect squares. Keep in mind, however, that after getting your answer you must check it so that the middle term is the correct one because it may not necessarily be a perfect square case just because the first and third terms are perfect square. The middle term also matters.
x² + 4x + 4
The following expression represents the factored form of a prefect square trinomial where a is the square root or the coefficient of the first term and b is the square root of the third term:
(ax + b)²
= (x + 2)²
Now to check our answer, we know it is correct if two times the product of a and b is equal to the middle term in the original trinomial. In this case two times 2 times 1 is equal to 4 which is the middle term so this factored form is correct.
y = (x+ 3)(x - 3)
(0) = (x+ 3)(x - 3)
Now to equal 0 we know that either the first multiple or second multiple has to be 0 (since 0 times any number is equal to 0) so we can set each of the multiples up to equal 0 and solve for x. After doing that you get the two zeros which are: (-3, 0) and (3, 0).
Next we find the vertex by first figuring out the AOS (Axis of Symmetry) which is in the exact middle of the two zeros. In this case it is 0. Now we set x to this number and solve for y to get the vertex:
y = ((0) + 3)((0) - 3)
y = (3)(-3)
y = -9
Therefore we know that the vertex is the point (0, -9). Remember that the vertex falls on the AOS so the AOS is its x value.
Now that you know what these points are you simply need to graph them. The following video shows a different example but one that has an a value. It also shows how to actually plot the points on a graph.
Q. A rectangle has an area defined by:
12x² + 17x + 6
Factor to find the algebraic expressions for the length and width of the rectangle. Then, if x represents 10 cm, determine the perimeter of the rectangle.
A. The first part of this questions is fairly simple as it is just asking you to factor the equation:
12x² + 17x + 6
= 12x² + 8x + 9x + 6
= 4x(3x + 2) + 3(3x + 2)
= (4x + 3)(3x + 2)
Therefore the length of the rectangle is (4x + 3) and the width of the rectangle is (3x + 2). For the second part of the question all we have to do is substitute in the length and width into the equation for the perimeter of a rectangle and then substitute in x as 10.
P = 2l + 2w
P = 2(4x + 3) + 2(3x + 2)
P = 2(4(10) + 3) + 2(3(10) + 2)
P = (86) + (64)
P = 150
Therefore, if x = 10, the perimeter of the rectangle is 150 cm.
Unit 3: Standard Form
- Complete the square to get vertex form.
- Solve using the quadratic formula.
- Use the quadratic formula to solve word problems.
y = ax² + bx + c
This form of the equation is not as useful as the other two because it doesn't reveal as much information about the quadratic equation. The value of c represents the y-intercept of the parabola but that is all we can conclude just from standard form.
In this unit we will be learning about how to complete the square to get vertex form. We already know how to factor to get factored form so it is natural we learn this as well. Converting the equation to vertex form gives us more information because it easily tells us what the vertex of the equation is which cannot be determined just from the standard form.
Also, we will be learning how to use the quadratic formula to solve the quadratic equation given the standard form. This is helpful because it allows us to easily figure out the zeros in one step instead of having to convert it to either factored or vertex form first,
Completing the Square
The following video provides an example of completing the square to get vertex form. The key points to remember are that a, which is the coefficient of the x² must be equal to one. If it is not, factor out the coefficient from the first to terms.
y = ax² + bx + c
The video below will further explain the quadratic formula and will provide an example of how to use it to solve for the two possible values of x. Remember their are two values of x because their is a plus or minus.
Q. The width of a rectangle is 6 meters less than its length and the area is 280 meters². Determine the dimensions of the rectangle.
A. First, let's declare our variable and draw a diagram:
Let x represent the length of the rectangle in meters.
A = xw
Keep in mind variable x here is representing the length. Now we can sub in the area and width to solve for x because we were given that the width is 6 meters shorter than the length:
(280) = x(x - 6)
Now that we have the base equation we can solve for x by putting it into standard form and using the quadratic formula:
280 = x² - 6x
0 = x² - 6x - 280
Substitute a, b and c into the quadratic formula to solve for x. In this case the values are: a = 1, b = -6 and c = -280:
x = (-b ± √(b² - 4ac)) / 2a
x = (6 ± √(36 + 1120)) / 2
x = (6 ± √(1156)) / 2
x = (6 ± 34) / 2
Now we have two continuations. We will try adding first:
x = (6 + 34) / 2
x = 40 / 2
x = 20
And now the second continuation in which we subtract instead:
x = (6 - 34) / 2
x = -28 / 2
x = -14
Clearly from the two results the value of x has to be 20. It cannot be -14 because a side length cannot be a negative number so -14 is invalid. Now that we know what the length is we can easily figure out what the width is because it is x - 6 which is 14 meters.
Therefore, the length of the rectangle is 20 meters and the width of the rectangle is 14 meters.
Also, any of the three forms of the equation can be used to graph the quadratic relation. When it is in vertex form you can easily graph the vertex and then use the step pattern to determine other points. When it is in factored form you can solve and plot the zeros. Then, you can use the axis of symmetry to solve for the vertex. When it is in standard form you can easily plot the y-intercept which is the third term and use the quadratic formula to solve and plot the zeros. Then you can plot the opposite point to the y-intercept and use the axis of symmetry to solve for the vertex.
The following example is one of the assessments I got back in which there were different possible ways to solve for the vertex and to graph the equation.
If you had used the quadratic formula to get the x-intercepts in question a) you could have subbed in the axis of symmetry to solve for the vertex instead. This just goes to shows how all of the different forms of the equation in a quadratic relation are related.