PAM - Creating Equivalent Expressions Lesson

Creating Equivalent Expressions

Techniques reviewed in this lesson will include factoring and completing the square. Changing forms from one form of a quadratic equation to another allows you to find different aspects of the graph.

Review

First, let's try a few ideas that we have already learned.

Try the distance formula and write an equivalent equation.

distance = rate * time may also be written as rate = distance / time

d = rt or when someone asks you to solve for the rate, $LaTeX: r=\frac{d}{t}$ $r=\frac{d}{t}$

Try the slope formula and write an equivalent equation, the point-slope formula.

The slope formula where m stands for slope and $LaTeX: \left(x_1,y_1\right)$ $\left(x_1,y_1\right)$ and $LaTeX: \left(x_2,y_2\right)$ $\left(x_2,y_2\right)$ are two points

$LaTeX: m=\frac{y_2-y_1}{x_2-x_1}$ $m=\frac{y_2-y_1}{x_2-x_1}$

Create the point-slope formula by clearing the denominator by multiplying $LaTeX: \left(x_1-x_2\right)$ $\left(x_1-x_2\right)$ to both sides.

$LaTeX: \left(y_1-y_2\right)=m\left(x_1-x_2\right)$ $\left(y_1-y_2\right)=m\left(x_1-x_2\right)$

In each case, the information is the same, but it is presented in a different form.

Notice that an equivalent form means that there is an equation, an equals sign between the two forms.

Equivalent Forms with Algebraic Expressions

Let's look at a simple example prior to using a more complex example.

An equivalent form for 1 yard is 3 feet, so 1 yard = 3 feet is the same information in a different format. These we looked at in the Ratios section in Data Analysis and Problem Solving.

For a fraction, an equivalent form for 3/2 is 1.5 or 1 ½. We could write 3/2 = 1.5 = 1 ½. Each provides equivalent information in a different form.

As we review more knowledge of math with different types of numbers and expressions in this module, we will continue to add information on equivalent expressions. A major portion of the SAT is being able to reason, to take given information and use it to solve a problem or create a solution that is reasonable for the situation described. Reasoning, using what you know, manipulating given information, is the key to these problems.

Convert from General Form to Vertex Form of a Quadratic Equation

To find the Vertex Form of the quadratic equation from the general form you must understand the parts of the general form equation and their relationship to the parts of the vertex equation or manipulate the general form equation into the vertex form by completing the square. Let's look at both methods.

Conversion to Vertex Form Using General Form a, b, c Relationships

Step 1: Find the axis of symmetry, which is, by the way, the x-value of the vertex.

Step 2: Plug the x-value of the axis of symmetry into the general form equation to get the y-value and the vertex is found.

Step 3: Begin writing the vertex form of the equation.

Step 4: Plug another easy number into the general form equation to find another point on the curve.

Step 5: Use the point found in the vertex equation to solve for the a value, the stretch.

Step 6: Write the vertex form equation.

Convert to vertex form

Step 1: Do you recognize the axis of symmetry from the equation? The axis of symmetry uses the formula , identified using the a, b, and c coefficients from of the general form equation

You have just found the x coordinate of the vertex. Remember the vertex is on the axis of symmetry as a parabola is symmetric.

Step 2: Now you can find the y-coordinate using the axis of symmetry by plugging the x from the axis of symmetry into the general form equation. Notice the function notation use for documentation and explanation.

f(1/2) = 1/2^2 - 1/2 - 2 Function notation to document (1/2) as input

Remember to get common denominators

Add/subtract the numerators only

You now have a point on the quadratic equation curve, the vertex ( , - ) clearly documented in the correct order from the last equation statement.

Step 3: Now you have found the vertex, so the vertex form can begin to be written. Vertex form is

Notice that only the stretch factor and direction of the parabola opening is not complete.

Step 4: To find a, plug in another easy number into the original general form equation. An easy number would be a number that makes little work, say 0 or 1. Let's use 0 since this is the number with less work.

Yes this one is easy and the point (0, -2) is easily seen from the function notation. The independent variable x (you picked) into the equation gives the dependent variable (answer based on x) y.

Step 5: Use the vertex form equation that you set up in step 3 with the new point from step 4 to solve for a.

With the new point (0, -2)

Plug in (0, -2) for (x, y)

-8 = a - 9 Multiply all terms by 4 to clear the denominator

1 = a Combine like terms

Step 6: Write the final vertex equation. Notice that this equation has fractions. Fractions are a part of the number system so we must practice using them.

Vertex is ( )

Stretch factor is 1

Positive coefficient 1 so upward parabola, so the vertex is a minimum point.

Note what the axis of symmetry, x = or vertex x-value tells you.

Decreasing graph -- < x ≤

Increasing graph < x <

Conversion to Vertex Form Using Completing the Square

This conversion form requires the technique of manipulating the general form equation using rules of math to create the square term of the vertex.

Convert to vertex form.

Step 1: Rewrite the general form equation using the x-intercept point of (x, 0) and solve for c.

In this case c = -2

Plug in (x, 0)

Solve for the constant number c

Step 2:

Whatever you do to one side of an equation, you do to the other

Using the coefficient of the middle term, -1, divide by 2 and then square the result.

Step 3: Now add the new number to both sides of the equation and create the square.

Add 1/4 to both sides

Rewrite the x variable side as a square. Use the -1/2 that you squared in step 1.

Simplify the left side

Move the constant to the x term side

Replace the (x, 0) point with (x, y)

You now have the vertex form equation (matches Example 1). The vertex is ( , - ) and the axis of symmetry is x =

Can't remember to change signs? Axis of symmetry is x = so take the portion of the equation inside the () and set = to zero and solve.

x - = 0

x = axis of symmetry

Convert from General Form

The equation must be factored. Either you can factor by hand, or you will need to use the quadratic formula, or graph and find the zeros. A graphing utility comes in handy with this.

The factors of the equation are the zeros, roots, the solutions of the equation when y = 0. The graph crosses the x-axis at these factor values.

Factoring with Knowledge of Multiplication

Step 1: Using multiplication, we can do this "by hand", without a graphing utility.

To factor by hand, you must determine two numbers that multiply to -2 and add to -1. To find these numbers, list the factors of the constant term and then see if two of them would add to -1.

The factors of -2 are in the table built below.

Factor 1	Factor 2
1	2	Need a negative sign to get -2
-1	2	What if 1 is - , then -1 + 2 = +1, no because we need -1
1	-2	What if 2 is -, then 1 + -2 = -1, got it!

We have found a solution.

We said that 1 and -2 are the numbers that work, so we can factor to get

$LaTeX: x^2-x-2=\left(x-2\right)\left(x+1\right)$ $x^2-x-2=\left(x-2\right)\left(x+1\right)$

Step 2: Write the factors in the intercept form equation.

y = a(x - b)(x - c)

y = a(x + 1)(x - 2)

Step 3: Find the coefficient a by choosing another easy point from the general form equation. Let's use the point (0, -2) again found in Example 1, step 4.

-2 = a(0 + 1)(0 - 2) Substitute in point (0, -2) for (x, y)

-2 = a(1)(- 2) Two negatives make a positive

-2 = -2a

1 = a Division by -2

Step 4: Finalize the intercept equation.

y = 1(x + 1)(x - 2) One times anything is the anything

y = (x + 1)(x - 2) Intercept Form

Factoring with the Quadratic Formula

Sometimes it is hard to factor the equations using the "by hand" multiplication method. The numbers do not lend themselves to quick factors. This is when the Quadratic Formula should be used. This formula always provides the x-values that are factors of the equation.

Quadratic Formula

Use with General Form y = ax² + bx +c

x =

where a, b, and c are the coefficients of the general form quadratic equation.

The SAT test will provide this formula on the formula sheet given so you do not have to memorize it. You need to know when and how to use it.

Convert to intercept form.

Step 1: From the general form equation given, a = 1, b = -1, and c = -2.

Step 2: Fill out and solve the quadratic formula and solve for both x's. Notice the ± sign. This sign says that and answer should be -b + the rest and -b - the rest.

- (-1) + or - square root of (-1 ^2 - 4(1)(-2) / 2(1)

Simplify terms:

Combine like terms

Simplify the radical if possible

Separate the two parts and find each x-value.

x = ( 1 + 3 ) / 2 x =( 1 - 3 ) / 2

x = 4 / 2 x = - 2 / 2

x = 2 x = -1

These values are the roots or zeros of the equation.

Make them into equation factors, the roots or anchors of the graph. The x-intercepts x = 2 and x = -1 is where the graph crosses the x-axis.

x = 2 x = -1

x - 2 = 0 x + 1 = 0

Step 3: Find the coefficient a by choosing another easy point from the general form equation. Let's use the point (0, -2) again.

-2 = a(0 + 1)(0 - 2) Substitute in point (0, -2) for (x, y)

-2 = a(1)(- 2) Two negatives make a positive

-2 = -2a

1 = a Division by -2

Step 4: Finalize the intercept equation.

y = 1(x + 1)(x - 2)

y = (x + 1)(x - 2) Intercept Form

From the intercept form the zeros, the roots, where the graph crosses the x-axis, the solutions are always where y = 0. These names are interchangeable.

So, 0 = (x + 1)(x - 2)

Simply take each factor and set the factor = 0 and solve for x.

(x + 1) = 0 (x - 2) = 0

x = -1 x = 2

Factoring Using a Graphing Utility

A graphing utility may be used to find the factors of the equation.

Step 1: Enter the equation in the y = section of the utility.

Step 2: Graph the equation.

Step 3: Find the zeros of the equation.

Step 4: Create the equation factors: x = a converts to x - a = 0

Step 5: Use another point on the curve to solve for the coefficient of a.

Step 6: Write the Intercept Form equation.

Convert from Vertex Form to General Form of a Quadratic Equation

Let's use the vertex form of the equation that we have been working with to show that we can go both ways.

Convert to General Form

Square the vertex term that is to be squared and combine like terms.

$LaTeX: y=x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}$ $y=x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}$

when added

Combine like terms - = -2

Convert from Vertex Form to Intercept Form of a Quadratic Equation

Convert to Intercept Form

Answer: y = (x + 1)(x - 2)

Convert from Intercept Form to General Form of a Quadratic Equation

Let's again use the same equation that we have been working with in intercept form.

y = (x + 1)(x - 2)

Foil the factors

Combine Like Terms

Convert from Intercept Form to Vertex Form of a Quadratic Equation

y = (x + 1)(x - 2)

Step 1: the x-coordinate of the vertex is the midpoint of the zeros.

Step 2: Convert to Vertex Form using the methods shown in General Form

Step 1: Evaluate the axis of symmetry from the factors

x = -1 and x = 2

( -1 + 2 )/ 2 = 1/2 x = equation image indicator is the axis of symmetry

Step 2: Complete Example 1, beginning with step 2.

Answer:

Notice that once you get familiar with the conversion methods, they are interchangeable based on what you understand from the equation given.

Are you curious by now what this equation that we have been working with looks like?

Here is the graph.

image graph x-intercepts at -1 and 2, the vertex, and the decreasing to increasing of the graph that changes at the axis of symmetry, x =

Notice the x-intercepts at -1 and 2, the vertex, and the decreasing to increasing of the graph that changes at the axis of symmetry, x =

Set x - = 0 and solve for x which is the axis of symmetry.

So, x = the axis of symmetry moves the graph to the right equation image indicator from the parent function or which has the vertex at (0, 0). So this graph is translated right and down from the parent function.

The domain for this graph is all real numbers as any x plugged into the equation will give a y-value. However, the range of the parabola equation is y ≥ - because the - is the lowest y-value on the graph.

Converting Vertex Form and Intercept Form Practice

Now it is your turn to try some conversions and answer some questions about situations involving these problems. Check out the handout Links to an external site. for detailed solutions.

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