Find the Quadratic Function That Models 3 Ordered Pairs

Find the Quadratic Function That Models 3 Ordered Pairs

Figure 1 is the solution to the second example on a post about finding the equation to a parabola when given three points.  The directions from the previous post read, “Find the quadratic function that models the given ordered pairs: (-2,-17), (-1,10), (5,-10).”

I plan on writing an explanation, but for now I wanted to get the solution up!

Cheers.



Find the Quadratic Function That Models 3 Ordered Pairs

Figure 1

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Modeling Data With Quadratic Functions

Find the Quadratic Function That Models 3 Ordered Pairs

In a previous post about quadratic functions and the graphs, I discussed the standard form of a quadratic equation, f(x)=ax^2+bx+c. Because a parabola is symmetric, its equation can be found by solving a system of equations with three variables. I have only discussed solving equations with 2 variables previously.  If you know how to solve a system with 2 variables, then a system with 3 variables should be no problem. The process is rather simple. I will explain how to solve the system of three equations as well as a system of two equations in the examples.

Example 1 Find the Quadratic Function That Models 3 Ordered Pairs

Before discussing how to find the solution, there should be a discussion about what is to be found. Ultimately, the solution should be an equation in the standard form of a quadratic, f(x)=ax^2+bx+c. To find this equation or function involves multiple parts and each part consists of several steps. Many times we get so caught up in the details, we forget the overall picture.

The answer is: f(x)=x^2-6x+7. Yes, I just gave you the answer. Now, the question is:

Find the quadratic function that models the given ordered pairs: (2,-1), (3,-2), (1,2). Here is how you turn 3 ordered pairs into a quadratic function.

First you must create a system of three equations. You create this system of equations by substituting the ordered pairs into the standard form of a quadratic equation f(x)=ax^2+bx+c. Using the reflexive and substitution properties of equality, I use the equation ax^2+bx+c=y. This will make solve the system of equations easier.

Substituting the ordered pairs gives the following equations:

(2,-1) gives a(2^2)+b(2)+c=-1

(3,-2) gives a(3^2)+b(3)+c=-2

(1,2) gives a(1^2)+b(1)+c=2

Simplifying these equations, maintaining the order from above and labeling them as equations A, B and C results with:

4a+2b+c=-1 (A)

9a+3b+c=-2 (B)

a+b+c=2 (C)

Now that you have a system of three equations, it is good to review the overall goal, to find the coefficients of a quadratic function that models the ordered pairs given. Okay, on with solving the system of 3 equations. You need to use the 3 equations and the elimination method to create a new system of equations with only 2 variables. You have to choose two pairs of equations and eliminate a variable. For this example, I will subtract C from A to create a new equation D and I will subtract C from B to create equation E.

(A) 4a+2b+c=-1

\underline{-(C) a+b+c=2}

(D) 3a+b=-3

(B) 9a+3b+c=-2

\underline{-(C) a+b+c=2}

(E) 8a+2b=-4

Equations D and E form a new system of equations.

3a+b=-3 (D)

8a+2b=-4 (E)

To solve this system, it would be easiest to subtract with E from equation D multiplied by 2.

2(3a+b=-3) (D)

\underline{-(8a+2b=-4)} (E)

6a+2b=-6 (D)

\underline{-(8a+2b=-4)} (E)

-2a=-2  Divide both sides by -2 and

a=1

You now have the coefficient of the quadratic term. Using a = 1, you substitute into either equation D or E. For this problem, I used equation D.

a=1

3a+b=-3

3(1)+b=-3  Substitute 1 for a.

3+b=-3  Simplify.

b=-6  Subtract 3 from each side.

This is the coefficient of the linear term. Now use both a = 1 and b = -6 to find the value of the constant. You must use one of the original equations A, B or C. I used equation C because it is very simple.

a=1 and b=-6

a+b+c=2 (C)

1+(-6)+c=2  Substitute 1 for a and -6 for b.

-5+c=2  Simplify.

c=7  Add 2 to both sides.

Finally, we have all of the coefficients a = 1, b = -6 and c = 7 and can write the quadratic function that models the given points.

a=1, b=-6, c=7

f(x)=ax^2+bx+c

f(x)=x^2-6x+7 This is the solution.

But how do you know it is correct? You should check your work by substituting the values a = 1, b = -6 and c = 7 into one of the equations A, B or C. Since I already used C I will use equation A.

a=1, b=-6, c=7

4a+2b+c=-1 (A)

4(1)+2(-6)+7=-1  Substitute.

4+(-12)+7=-1  Simplify.

-8+7=-1  Simplify.

-1=-1  Simplify and it checks.

This is just one example of finding the equation of a quadratic function from three points. Also, because the three original equations started with all three variables, it was a more complicated type.

Example Find the Quadratic Function That Models 3 Ordered Pairs For You To Do

Find the quadratic function that models the given ordered pairs:

(-2,-17), (-1,10), (5,-10).

Click here for the solution.

Quadratic Functions and Their Graphs

Standard Form of a Quadratic Function

f(x)=ax^2+bx+c

Where a\neq0.

ax^2--> Quadratic Term

bx--> Linear Term

c--> Constant Term

The restriction on the quadratic term, a\neq0., ensures that the function is indeed a quadratic function. Without the squared term, you are left with a linear function or a constant function.

A simple skill assessed in most high school algebra 2 classes is the ability to classify functions. If you are studying quadratics, then you have studied linear equations with their close relative the constant function. To classify a function, it is often necessary to simplify what is given in order to properly determine the type of function.

Example 1 Identifying Functions

Classify each equation as linear or quadratic and identify the quadratic, linear and constant terms.
y=(4x+2)(-3x-1) (1)

y=(4x)(-3x) + (4x)(-1) + 2(-3x) + 2(-1) (2)

y=-12x^2 + (-4x) + (-6x) + (-2) (3)

y=-12x^2-10x-2 (4)

In the steps above, line 1 represents the given function. Line 2 represents simplifying the equation with multiplication. You may have heard of the acronym FOIL, which stands for: First, Outer, Inner, Last. This is a short cut to multiplying to binomials. Anyway, line 3 is formed by actually performing the multiplication set up in line 2. The change from line 3 to 4 occurs by collecting or adding the like terms of -4x and -6x to give the linear term -10x. The quadratic term is -12x^2 and the constant term is -2.

Example 2 Understanding the Graph of Parabola

As with working with linear functions, the ability to interpret and analyze the graphs of basic quadratic functions is critical. It is good to note that the graph of a quadratic function is called a parabola. Figure 1 is the graph to the parabola, f(x)=x^2+8x+14. For now, the function in the previous sentence is irrelevant. The reason for this, the graph is given. I will discuss in an upcoming post how the actual function is related to the graph of the related parabola.

Firgure 1 Parabola

Figure 1 Parts of a Parabola

In figure 1, f(x)=x^2+8x+14 is graphed and has vertex of (-4,-2) and an axis of symmetry (AOS) of x=-4. Though I did not label the other two distinguishable points, you should be able to see, that the other points on parabola correspond to each other. This is because parabolas are symmetric, hence the axis of symmetry. Those points are equidistant from x=-4. Since this parabola opens upward, it has a minimum or bottom value. You get the value from the y-coordinate. The reason you use the y-coordinate is there is not bound on the Domain. Parabolas extend in the both directions along the x-axis. Because of the shape of parabola, there is a limit on the range. In this case, the range could be expressed as {y|y≥-4}.

The previous paragraph can be boiled down to identifying the AOS, vertex, points and corresponding points on the graph of a quadratic function or parabola.

Figure 2 Parts of a Parabola

Figure 2 Parts of a Parabola

Example 3 Understanding the Graph of Parabola For you to do.

For figure 2, identify the axis of symmetry, the vertex, the minimum or maximum value and explain why it is a min or a max value, identify both points and their mirror image about the AOS.

Click here for the solution.

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