Download LESSON #36: GRAPHING QUADRATIC FUNCTIONS

Graphing Quadratic Functions
Lesson 36
Your warm-up for lesson 36 has your students practice factoring polynomials. This is
a very important skill because they are going to have to factor trinomials in lesson 36
in order to find the x intercept. If you were to graph the quadratic equation y=x2-x+2,
what would be the axis of symmetry and vertex? If we “S” this problem, or studied
this problem, we would first begin by underlining the question. What would be the
axis of symmetry and vertex? We will also answer the question, what is this problem
asking me to find? This problem is asking me to find the axis of symmetry and
vertex. In problem one, we will first complete the table in order to find the
coordinates to graph the quadratic equation. This is the same as any other function
table, where our x value is our input and our y value is our output. We will input -4
into the function to order to find our output. y=-(4)2+3(-4)+2. To simplify this
expression, we will simplify the exponents. -42=16; we will then multiply 3(-4)=-12;
and then we will add 16+ -12+2= 6. So our coordinate is -4,6…this is one of the
points on the quadratic equation. We can also plug in -3. Notice you are plugging
the input in for each x in the equation. If is very important that you use parentheses,
so your students will know your students will know they need to multiply (-3) (-3).
It’s very important when you are working with integers and the negatives have to
cancel. (-3)(-3)=+9; we will add our 3 (-3), which gives us -9; +2; so this leaves us with
the point -3,2. If we plug in -2… (-2)2 =4; + 3(-2)=-6; +2; 4+ -6+2= 0. You will
continue to plug all of your inputs in to find all of the outputs. To graph problem one,
after you have created your table with your inputs and outputs, you will then graph
each of the coordinates. Our first coordinate is -4,6. We will begin at 0,0…move to
the left 1-2-3-4 units and move up 6 units and make a point. Our second point is 3,2. We will start at 0,0 and move to the left 3 and up 2; -2,0; -1.5.-0.25; -1,0; 0,2;
and 1,6. When your students graph these points they should notice that they are not
in a line; quadratic equations create a graph, in the shape of a u, and we call this a
parabola. They will connect their points in order to draw a parabola.
In problem 2, we will plug in our x coordinates in our input into the quadratic
equation for x; and then simplify to get our output y. If is very important that you use
parentheses when you are plugging in. Notice that we have a –x2, so the negative
stays on the outside, and you input a -4 in parentheses, which will be squared; minus
2 (-4)+3. When you square -4… -4(-4) it gives you a +16; however with the negative
out here, it stays as -16; we then have a negative two or -2(-4) which is +8 +3. 16+8+3, leaves us with -5. To find our next y coordinate, we will plug in -3 for x,
leaving your negative on the outside. We have -32… negative times a negative is a
positive, so this gives us +9 but then it’s multiplied by -1, which gives us -1 on the
outside so it’s -9; -2(-3)=+6+3.
-9+6+3 leaves us with a sum of zero. So -3,0
is the coordinate for our quadratic equation. And also plug in -2. -2(-2)=4 and then
when we multiply by the negative, it because a -4; -2(-2)=+4; +3; -4+4+3…leaves us
with an output of 3. You will continue to use this process with the other inputs in
order to find the outputs. If we were to graph, y=-x2-2x+3, we will use our table that
we have created. To graph the point -4,-5… we will begin at 0,0 and move to the left
4 units and then move down 5 units and make a point. -3,0…. You will move to the
left 3 units and we will not move up or down at all. -2,3…we will move to the left
two units and then move up 1-2-3 units. -1,4…one unit to the left and then up four
units. 0,3…we will not move to the right or left at all, we’ll move up three units. 1,0.
And then, 2, -5. We see once again, that our points do not make a line, they make a
“U” shape. Students can connect the dots, and see the graph of their quadratic
equation, which is called a parabola. When we look at our two problems which we
have graphed, we know that these are quadratic equations because they are in the
form y=ax2+bx+c. Both of them have an x2 term and an x and a consonant. Both
graphs create what is called a parabola, or a u shaped graph. They do not go in a
linear, straight line. When you have an ax2 or an x2 term, this is going to create your
parabola, as long as the x2 is the largest variable that you have. If you notice our a in
problem 1 is a 1 or a +1…anytime our a is positive our graph is going to open
upward, creating a minimum point at the bottom. This minimum or point at the
bottom is called a vertex. If you look at problem 2, we see the coefficient of the x
term is -1, and when this is negative, our parabola opens downward. It creates a
maximum point, or a higher point, which is still called a vertex. Another property that
all parabolas possess, is symmetry. If you fold them over, they are symmetrical—
there is a line that you can use where the graph will be the same on each side. We
call the line of symmetry, the line that cuts down through the middle. And your vertex
is always on your line of symmetry.
For problem one, if we want to find the line of symmetry, we can use this formula—
x= -b/2a (-b divided by 2a). This will always give you your line of symmetry. And
because x equals a number, is a vertical line; it is the equation of a line. We said that
the equation was in the form, y=ax2+bx+c. We can identify our a, our b, and our c.
The coefficient of the x2 term, or a, is 1; the coefficient of the x term, or b, is 3; and
our constant is 2. If we plug into the equation, x= -b, or -3 over 2(a), in this case we
have 1 as our a…so x=-3/2. Some students may want to write this as a decimal,
which would be -1.5 Because the vertex is always on the line of symmetry, this
means that our x coordinate will also be a -1.5 for the vertex. Now that we know our
x coordinate using the line of symmetry, we can plug back in to the original equation
in order to find the y coordinate. Write your original equation and plug in -1.5; we will
simplify our exponents: a negative times a negative is a positive; we will simplify our
multiplication, -4.5(2); and when we simplify, it leaves us with y=-0.025. Okay, now
we have found our axis of symmetry and our vertex. In problem 2, to find our axis of
symmetry, we will first identify our a, b, and c; we know that our equation is in the
form of y=ax2+by+c. a equals, in this case, the coefficient of x2, which is -1; b= -2;
c=3. We plug this in. x=
-(-2) over 2 (-1). –(-2)= +2; and 2 (-1)= -2; 2 divided by -2
is -1…so, x= -1. Our axis of symmetry is x=-1. We know that the vertex is always on
the axis of symmetry. So, if x=
-1, there’s -1 in our vertex. And we can plug this back in to our original equation in
order to find our y coordinate. When we multiply (-1) (-1), it leaves us with a +1;
however, we then multiply it by the negative in front…so we have y= -1; a negative
times a negative is a positive 2; +3; -1+2+3, leaves us with 4. And our vertex is -1,4.
In order to find your x intercept, you must factor this trinomial. We are in the form
y=mx2+bx+c; so if we have a=1 and b as 2; a(b)= 2 and b= 3. We need two numbers
that multiply to 2 and add to 3. So, 1 and 2: 1(2)=2, 1+2=3. We have x2+3x+2. We
know that 1+2=3; so 1x+2x…has got to be equal to 3x. This will allow us to split our
middle term into 1x, or x, and 2x. If we group the first 2 and the last 2. The greatest
common factor of the first two terms is x; x2 divided by x= x; and x divided by x is 1.
The greatest common factor of our last two terms is 2; 2x divided by 2 is x; and 2
divided by 2 is 1. x+1 is a factor they share. x+2 is what’s left over. And we can set
both of these binomials equal to 0. x+1=0; x+2=0. We can solve for x after we factor
because we only have one x. If we have a trinomial we cannot solve for x because if
you were to subtract 2, you’d still have and x2 and an x, which would be impossible to
solve for x. We now can solve for x by subtracting 1 from both sides, leaving us with a
-1; and subtract 2 from both sides, and x=-2. We factor to find your x intercept. And
we can check this by going back to our graph. We intercepted the x axis as -2 and -1;
and x=-1 and -2. We also know that our y intercept is going to be our c because if we
were to plug in 0, which is always the x coordinate of our y intercept, then we would
be left with 2. So, c= our y intercept…so our y intercept in this case is = 2. If we go
back to our graph, we can see that the parabola does intersect the y axis at 2; it
crosses the axis at 2. So we can factor our trinomial and set both of those binomials
equal to zero to find our x intercept; and we can use our c as our y intercept.
In problem 2, we have y= -x2-2x+3. And we are going to factor the trinomial. We are
in the standard for of y=ax2+bx+c. Our a=-1; our b=-2; and our c=3. Our a(c)=-3;
and our b=-2. So we need two numbers that multiply to be -3 and add to be -2. -3
and 1 will work. We know that -3+1=-2, so -3x+1x=-2x. This allows us to split our
middle term.
-2x=-3x+1x +3. If we group our first two terms and group our last two terms. Some
students may want to use the box method, at this point. Our greatest common factor
is
–x. –x2 divided by –x; and -3x divided by –x is +3. Our greatest common factor in the
second binomial is just 1, so we are left with x+3. One of our factors is x+3, what
they share. And our other factor is –x+1, or what is left over. We will set both of
these binomials equal to zero, in order to solve for x. Subtract 3 from the both sides.
Subtract 1; divide by -1. We have found that our x intercepts are at -3 and 1. We
also said that our c will be our y intercept. So our y intercept should be at 3. If we
look back at our graph, our x intercepts are at -3 and 1 and our y intercept is at 3.
Our y intercept is supposed to be at 3; and our x intercepts are -3 and 1. To make
our 8 page booklet foldable for quadratics, we will first begin by making a “hot dog”
fold, folding the paper vertically. We will then hamburger fold, and hamburger fold
again. We open this up and you are going to find the double folded section, this part
right here, it’s folded on this side and folded here. And you are going to trim, just
barely trim the top of that off to the middle crease. Don’t trim all the way out…trim
from that double folded section over to the middle. We then open it up, and slide
your fingers here, and pull out—we create a star, and then actually just fold that
down. Create your little booklet and you can write quadratics on the front and your
examples with finding the x intercept and graph and your vertex inside.
We have already “S” the problem, so we know that this problem is asking me to find
the axis of symmetry and the vertex. In “O” organize the facts, we are going to read
the problem and decide which facts are necessary and unnecessary. If you were to
graph the quadratic equation y=x2-x+2, that’s our fact and it is necessary so we will
write it down. In “L” line up a plan, we will choose our operation or operations and
then we will write in words what our plan of action will be. Because we are going to
find the line of axis of symmetry and the vertex, we will be using division and
multiplication. We will also use addition to simplify. We will write in words what our
plan of action will be. I will use the axis of symmetry formula and substitute the x
value into the equation to find the y. When we use the formula, we will find the axis
of symmetry. And when we have x and y, we will then have our coordinates for the
vertex. In “V” verify your plan with action, we will first estimate our answer. Because
our equation is y=x2-x+2, all of the variables: a, b, and c are very small numbers. So
we can estimate that our axis of symmetry and our vertex will have coordinates which
are less than 10. To carry out our plan, we will first find our axis of symmetry. You
know that the axis of symmetry is equal to –b divided by 2a. If we look at our original
equation, y=x2-x+2. a=1; b=-1; and c=2. So we are going to have a negative -1 over
2(1); -1(-1) is a positive 1; and 2(1)= +2. So we have ½ . So our x coordinate is going
to be ½ ; and our axis of symmetry is going to be x= ½. And then to find our y
coordinate of the vertex, we will plug back in. y=x2-x+2; y= (1/2)2
-(1/2) +2. ½
squared is ½ times ½ or ¼ - ½ +2; and when we simplify, y= -9/4. This gives us a
vertex of ½,-9/4. In “E” examine your results, we will first ask ourselves, does your
answer make sense? We will go back to our question, or the problem was asking us
to find …the line of symmetry and the vertex. And we do have an equation which
represents the axis of symmetry, and we do have a coordinate which represents the
vertex. So our answer does make sense. Our second question is, Is your answer
reasonable? If we look back at our estimate, we said the coordinates would be less
than 10, and everything is less than 10 so our answer is reasonable. And, Is your
answer accurate? You can have your students rework the problem, they can plug the
coordinate of the vertex in and they can make sure that it makes a true statement.
They could graph the parabola and see if the axis of symmetry is at ½ and if it has a
vertex of (-1/2, 9/4). And then we are going to write our answer as a complete
sentence. The axis of symmetry for the quadratic equation y=x2-x-2 is x at ½ and the
vertex is the coordinate pair (1/2,
-9/4). We are going to close our lesson by
looking at the essential questions. Number 1, what is a quadratic equation? A
quadratic equation is a equation with the basic form y=ax2+bx+c, whose graph is a
parabola. Number 2, How do we graph a quadratic equation? You can use the axis
of symmetry, the vertex, the x and y intercepts, or fine points to graph a quadratic
equation.
In lesson 36, we can graph our quadratic equation on our graphing calculator. We will
press the y= button and clear out anything that we already have. We will enter our
equation, y=y2+3x+2. From here you can go to the window and see but if it’s on the
standard window, the x axis goes from -10 to +10, with a scale of 1 and the y axis
goes from -10 to +10, with a scale of 1. When you press graph, you will see your
parabola. You may need to change the window from time to time depending on your
equation of the parabola. In order to find the maximum or the minimum points, you
will press second trace for calculate. You see that you can choose, either three is the
minimum or 4 is the maximum. Looking back at our graph, this parabola has a
minimum. So when we press second calc, we will choose #3 for minimum. Looking
at the bottom of the screen, it asks you to find the left bound of the minimum? This
is very important. You must use your arrows and move the cursor to the left of the
minimum. If you don’t, you may not get the correct minimum points. Once you have
moved the cursor to the left, you press enter. Then the bottom of the screen asks you
to move to the right of the minimum. When you have moved your cursor to the right,
you press enter again. It then asks if you want to guess and you say yes by pressing
enter. And we can see that the minimum point or the vertex of this parabola, is that 1.5, -0.25. Any time you have numbers that go on and on and on with zeros, you can
round that off. If we were going to write our minimum, it would be (-1.5, -0.25). We
can also find our x intercept. In order to do this, you will press second trace again.
And looking at the calculator, we are going to find our zero. A zero is another name
for an x intercept. You can either arrow down and press enter on two or press two.
Once again, it is very important that you move your cursor. There are two zeros for
this parabola, so it is very important that you move to the left if you are finding the far
left zero and press enter; and then move to the right: of the far left, you can’t go all
the way over here because that will be to the far right of both of the zeros, you have
to go to the right bound and to the left bound of one zero, staying away from the
second zero. Once you press enter, you press guess, press guess again…and it tells
us we have a zero at -2. This means the parabola crosses the x axis at -2. One of our
x intercepts is -2. If we go back to the calculator again, and press second calc, we
are going to find our second zero by pressing 2, it asks us to find the left bound, so if
you’re finding the far right zero, you’re okay because you are to the left of the second
zero. And now you can move your cursor all the way to the right, to the right bound,
press enter again for guess and our second zero is at -1. So our two x intercepts are
(-2,-1).