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## Using Properties of Parabolas to Graph a Parabola Problem 2

As I discussed in an earlier post on graphing a parabola, the following information allows to easily graph a parabola in standard form $f(x)=ax^2+bx+c$.

Properties of a Parabola or a Quadratic Function in Standard Form

Given that $f(x)=ax^2+bx+c$ is a parabola when $a\neq0.$

1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
2. The axis of symmetry is given by the line $-\frac{b}{2a}$.
3. The vertex is given by the ordered pair $(-\frac{b}{2a},f(-\frac{b}{2a}))$
4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

In my first post the equation did not have a linear term and was in the form $f(x)=ax^2+ c$. The vertex could be found using (0,c) and the axis of symmetry (AOS) of x = 0. In terms of the coordinate plane, the y-axis is the AOS.

In this post, we change gears and will be graphing a parabola that is given in standard form $f(x)=ax^2+bx+c$. The point of this work is to sketch a graph using the core concepts behind the parabolic relation. Remember, because a parabola is symmetric every point except the vertex has a mirror image. Once you find the AOS and vertex, you will have to use the same process to find another point and then its mirror image. Let’s take this to the example.

Example Graphing a Quadratic Equation in Standard Form

Graph $f(x)=2x^2+4x-3$

Looking at the properties above in order, first we know this graph will open upwards, this is due to the fact the coefficient on the squared term is positive or a>0. The AOS is found by using the ratio $-\frac{b}{2a}=-\frac{4}{2(2)}=-1$ and can be written as x = -1. If you are trying to create the graph at the bottom of this post, you should draw a vertical dashed line at x = -1, because this represent the AOS.

Next we find the vertex. Since the equation or function is given as $f(x)=2x^2+4x-3$ and x = -1, it is natural to substitute -1 to get $f(-1)=2(-1)x^2+4(-1)-3$. Using the order of operations to simplify the right side, we arrive at f(-1)=-5. If you have no idea what is going on, here is an explanation. The AOS serves as the x-coordinate of the vertex. Since we know the x-value, we can find the y-value (***f(x) is the same as using the variable y because of ‘function notation’***). By plugging in the known x-value, we can find the related y-value and we now have an ordered pair. For this example the vertex is the ordered pair is (-1,-5). If you are trying to graph this problem, you should plot the point (-1,-5)

Lastly, we need to find two more points. To do this, pick any x-value to the left or right of the AOS. For this example I will use x = -3. Using the given function, $f(x)=2x^2+4x-3$, substituting -3 in for x gives $f(-3)=2(-3)^2+4(-3)-3$ applying the order of operations f(x)=3 and the ordered pair (-3,3). Again, because parabolas are symmetric, (-3,3) has a mirror image at (1,3). If you are graphing along with the post, plot the points and draw a curve to fit those points. Even though this is a sketch, you should try to make it neat.

Graphing a Parabola in Standard Form

## Using Properties of Parabolas to Graph a Parabola

When graphing a parabola it is good to know that it is symmetric. A parabola will have a vertex. The vertex is the minimum or maximum value of the parabola. It is good to note, the range has a restriction on it, not the domain. The vertex represents the axis of symmetry as I discussed in a post about modeling data with quadratic functions.

Starting with the standard form of a parabola $f(x)=ax^2+bx+c$, you can determine the where the vertex will lie in the coordinate plane and other aspects of a parabola. Below are the properties of a parabola or a quadratic function in standard form.

Given that $f(x)=ax^2+bx+c$ is a parabola when $a\neq0.$

1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
2. The axis of symmetry is given by the line $-\frac{b}{2a}$.
3. The vertex is given by the ordered pair $(-\frac{b}{2a},f(-\frac{b}{2a}))$
4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

The above information makes sketching the graph and finding the actual value of the maximum and minimum value of a parabola very easy. Just as you can write the equation of a parabola from three points, it is important to graph at least three points for a parabola.

Graph $-\frac{1}{2}x^2+2$. To graph any quadratic it is important to get an idea of what it should look like from the properties. In this case a < 0. Which means the parabola, will open downward. Often, parabolas are referred to as u-shaped graphs, so the u will open downwards. The axis of symmetry (AOS) will be at

$-\frac{b}{2a}=-\frac{0}{2*-\frac{1}{2}}=0$.

This function does not have a linear term, therefore the AOS is given by x = 0. Any parabola without a linear term will have an AOS of x = 0. Finding the vertex requires you to substitute 0 in for the variable x. This is where function notation comes in handy,

$f(0)=-\frac{1}{2}0^2+2$

$f(0)=2$.

Thus, the vertex $(-\frac{b}{2a},f(-\frac{b}{2a}))=(0,2)$

Because parabolas are symmetric, you need to only find one point other than the vertex. The third point is found by finding the corresponding point of the second point found. In this case, you would want to find f(2), which means when x = 2. The related y-value would be found with the following steps.

$f(2)=-\frac{1}{2}(2)^2+2$  substitute 2 in for x

$f(2)=-\frac{1}{2}(4)+2$  Square the 2

$f(2)=-2+2$  Multiply

$f(2)=0$  Add

Figure 1 Parabola y = (-1/2)x^2+2

﻿This gives a point of (2,0). The point the is symmetric about the AOS is (-2,0).All of the key points and the AOS are mark in Image 1 below. It is good to be worried about neatness when creating a graph, but if you are doing it by hand a good sketch will do. To the left is the best I could do using photo shop. If you need a an accurate graph, then using  graphing calculator or computer software would be your best bet.

## 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.

﻿

Figure 1

## 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).

## 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.

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

### 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.

## Parabolas – Identifing Parts of a Parabola

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

Identifying Parts of a Parabola

For the figure, 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.

The axis of symmetry is $x=-\frac{3}{2}=-1.5$. Since the parabola is facing up the y-coordinate of the vertex represent a minimum value. The vertex is $( \frac{3}{2}, \frac{5}{2} )$. The minimum value is $\frac{5}{2}$.
Point A is $( -\frac{5}{2}, \frac{9}{2} )= (-2.5,4.5)$ and its corresponding point would be $( -\frac{1}{2}, \frac{9}{2} )= (-0.5, 4.5).$
Point B is $(0, 7)$ and its corresponding point would be $(-3, 7).$