PAM - Creating and Solving Quadratic and Exponential Equations Lesson

 

Creating and Solving Quadratic and Exponential Equations

In the Heart of Algebra Module you created the equations of lines from a point and a slope, two points, or a graph.  We used the y-intercept form or the point-slope form of the linear equation formulas to create the equation.  Do you remember the two formulas and how to use them with the slope formula when needed?

Creation of Line Equations

Using the Line Formula Review Flash Card Activity below, see if you can remember the appropriate formula and the information needed in the formula prior to flipping the card.   

 

The creation of quadratic and exponential functions also may be created from points, using appropriate formulas for these functions. Let's start with the creation of quadratic equations and then we will move on to exponential equations.

Creation of Quadratic Equations

In the Problem Solving and Data Analysis Module, we examined briefly a few of the formulas that a parabola (quadratic equation) could be identified with.   The general form and the vertex form were examined detailing what can be identified from the forms.

Intercept form of a quadratic equation is an additional form that we will study here. This form provides the x-intercepts clearly. The x-intercepts are the x-values where the parabola crosses the x-axis.  

So, let's create some quick notes on the types of quadratic forms that will be used to create a quadratic equation. Note that y may also be written in function form, f(x).  Both mean the same thing, y = f(x).  The f(x) notation allows you to indicate which x-value is being placed in the equation to find the y-value.

An additional form that will be examined here is the intercept form.  The quadratic equation intercept form means that we are working with the x-intercepts; where the parabola crosses the x-axis. Please do not confuse this with the y-intercept form of the linear equation. You will note that the y-intercept form clearly indicates the axis as y.

Now let's examine the thought process to decide how many equations are needed to find missing parts of an equation.

• The point-slope formula is used with two points to find the linear equation. Remember that the highest exponent of x is 1 for a linear equation.
• So how many points would you think we would need to find a quadratic equation? Remember that the highest exponent of x is 2 for a quadratic equation.
• Did you choose 3? If you did you are beginning to see a pattern. You need one more point than the highest exponent to create an equation.

Another pattern is found in solving a system of equations. When you solved for the intersection of two linear equations you had to find the specific x and y where the equations intersected.  You had two unknowns (the x and y) so you needed to have two equations to solve for them.  We used substitution, elimination, or graphing to solve these equations.  Review the Heart of Algebra, Systems of Linear Equations in Two Variables lesson.

Quadratic Equations Creation Using General Form

Example 1:  Given points (−3, 9), (0, 0), (2, 4), find the quadratic equation that will contain all three of the points in standard form.

We know that the general form of a quadratic equation is y = ax² + bx + c.  We also know the values of x and y because we were given points, which are relationships in x and y. So what is missing?  We don't know a, b, and c.  To find the equation we will need to find a, b, and c.  

Step 1:  Create three equations with the three points using the general form.

9 = a(−3)² + b(−3) + c                                    

9 = 9a          − 3b     + c                            

 

0 = a(0) + b(0) + c

0 = 0a    + 0b   + c    

 

4 = a(2) + b(2) + c

4 = 2a    +2b    + c

 

0 = c

Notice that the middle equation with point (0, 0) worked out to provide an answer for c.  When this happens, c can be filled in for the other equations.  So now we have two equations and two unknowns; a and b are the unknown variables.  

You have discovered the pattern for simultaneous equations. The number of unknowns is the number of equations you must have in order to find the solution, the intersection of information.

Step 2:   Using the concept of Simultaneous Equations to solve this set of two equations for the intersection point, the point which will work in both equations.

The two remaining equations are:

9 = 9a − 3b + 0      

4 = 4a + 2b + 0

Notice that the first equation may be reduced by dividing everything by 3 to make it easier.  

3 = 3a − b    12 = 12a − 4b

4 = 4a + 2b −12 = −12a − 6b    

Note that we remembered systems of equations and multiplied each equation by a number that would allow one of the variables to add to zero, allowing for the solution of another.

Step 3: Solve for the last variable using one of the equations.  

3 = 3a - 0.   With division by 3, a = 1.

Step 4:   Plug the solutions for coefficients a, b, and c into the general form to find the equation.

So the equation that contains the 3 points given on its curve is

y = 1x² + (0)x + 0 or y = x².

Note that all of the quadratic equations do not necessarily have a, b, or c as zero. This is a special case.

Watch the video to see another example of solving for an equation that has three unknowns.  

Now that you have seen two quadratic equations solved from three points, it is your turn to try. See the General Form Practice Solutions Handout Links to an external site. for more detailed information.

Quadratic Equations Creation Using Vertex Form

Example 2:   Given a vertex (−1, 2) and another point (4, 6) write the quadratic equation for the parabola.

Since the vertex is given it is quick to use the vertex form.

Remembering from the Graphs of Linear, Quadratic, and Exponential Functions lesson of Problem Solving and Data Analysis, the vertex is symbolically

(h, k).      

Step 1:  Using y = a(x − h)² + k 

y = a(x-(-1))² + 2

y = a(x + 1)² + 2

Step 2:  Find a by plugging in the other point for x and y and solving for a.

6 = a(4 + 1)² + 2

4 = a(5)²

4 = 25a

4/25 = a

Note that a could be any real number that includes fractions and decimals.

Step 3:  Rewrite the equation in vertex form inserting a and leaving x and y.

$y=\frac{4}{25}\left(x-2\right)^2+2$  or   y = .16(x − 2)² + 2

Watch the video and you will see another example of using the vertex form of the equation. This example will begin with a graph. You will be required to determine the information that you need from the graph provided.

  Now it is your turn to try placing quadratic equations in vertex form. See the Vertex Form Practice Solutions Handout Links to an external site. for more detailed information.

 

Quadratic Equations Creation Using Intercept Form

x-Intercepts

The intercept form for parabola equations uses the x-intercepts, not the y-intercepts.   The x-intercepts are called the solutions to the equation because these have y-values of zero.  

Example 3:  Given the x-intercepts −2 and 4, find the quadratic equation that passes through the point (5, −1).

Step 1.   Set x= to each of the intercepts and then move the intercepted number to the same side of the equation as x.

x = −2                             x = 4

x + 2 = 0                         x − 4 = 0

These are the factors of the quadratic equation.

Step 2. Write the intercept equation using the factors found in step 1.

y = a(x+ 2)(x − 4)      

Step 3.   Solve for a using the given point (5, −1) that the equation must pass through.

−1 = a(5 + 2)( 5 − 4)

−1 = a(7)( 1)

−1 = 7a

−1/7 = a

Step 4. Rewrite the equation in intercept form with the a-value and factors.

 $y=-\frac{1}{7}\left(x+2\right)\left(x-4\right)$

Leave a in fraction form if the denominator is not divisible evenly into the numerator. This is the best form as this is an exact answer.

However, at times the SAT test may approximate. Pick the closest answer.

Watch the Intercept Form video to learn to find the information from a graph and create an intercept form equation.

  Now that you have watched the video for another example, try a few problems on your own. See the Intercept Form Practice Solutions Handout Links to an external site. for more detailed information.

 

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