1. Previously, we worked with linear functions, in which the highest exponent of the variable x was 1. In this section, we find ourselves working with quadratic functions, for which the highest exponent of the variable x is 2.
Definition Any function that can be written in the form f(x)=ax2+bx+c, where a, b, and c are all real numbers, is called a quadratic function. The graph of a quadratic function is a parabola. x y 8 6 4 2 -2 -4 -6
Figure (a)
When the value of a (from f(x)=ax2+bx+c) is a positive number, the parabola will open upward.
f(x)=x2+3 is shown in figure (a). For this function, note that a=1, a positive value. So the parabola will open upward. x y 8 6 4 2 -2 -4 -6
Figure (b)
When the value of a is a negative number, the parabola will open downward.
f(x)= -x2-2x+3 is shown in figure (b). Here, a=-1, a negative value. So the parabola will open downward.
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2. The vertex is the point (h,k).
If a parabola opens upward, the vertex’s y-value will be the smallest number in the function’s range.
The parabola graphed in figure (a) has its vertex at (0,3). This function’s smallest value in its range is 3. x y 8 6 4 2 -2 -4 -6
Figure (a)
Vertex:(0,3)
A very important ordered pair that should be identified on every parabola is the vertex.
If a parabola opens downward, the vertex’s y-value will be the largest number in the function’s range.
The parabola graphed in figure (b) has its vertex at (-1,4). This function’s largest value in its range is 4. x y 8 6 4 2 -2 -4 -6
Figure (b)
Vertex:(-1,4)
The vertex is the point (h,k).
Every parabola is symmetric with respect to a line that contains the vertex. The equation for the parabola’s line of symmetry is x=h.
If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward.
If |a| > 1, the parabola is narrower than the basic parabola.
If |a| < 1, the parabola is wider than the basic parabola

3. Example 1. Given the function f(x)= 2(x  3)2 7 , find the following: Vertex, line of symmetry, opens up or down, wider or narrower than the basic parabola.
Solution:
h=3; k=-7, so the vertex is (3,-7).
Then, the line of symmetry would be the equation is x=3.
Since a = 2, the parabola opens upward and is narrower than the basic parabola.
Your Turn Problem #1

4. Vertical Translations
The basic parabola is the functions f(x) = x2.
Let’s make a table and find some ordered pairs. Since h = 0 in this equation, pick two points to the left of 0 and two points to the right of 0. x y 4 3 2 1
f(x)=x2 ●
x y
0 0
1 1
2 4
We can see the graph is symmetric with respect to the line x = 0, (the y-axis).
A graph of a function of the form f(x)= x2+k, where k is a real number, vertically translates the graph of the function f(x)= x2 up k units (if k is positive) or down k units (if k is negative). x y 5 4 3 2 1
f(x)=x2
g(x)=x2+3 6 7
Example 2. Graph g(x)=x2+3.
Solution:
This graph has the same shape as the graph of f(x)= x2, but every point has been raised 3 units.
The graph of f(x)=x2 has been vertically translated up 3 units to give the graph of g(x)= x2+3.
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5. x 4 3 2 1
f(x)=x2
f(x)=x2-4 -1 -2 -3 -4 y 5 6 7
On homework or a test, only the requested function needs to be graphed. If you wish to have both the basic parabola and the translated parabola on the same grid, each parabola should be appropriately labeled.
Vertically translate every point on the graph of f(x)= x2 down 4 units (since k=-4).
Your Turn Problem #2

6. Horizontal Translations
Example 3.
Graph f(x)=(x  1)2.
Solution:
In general, a graph of a function of the form f(x)= (x - h)2, where h is a real number, horizontally translates the graph of the function f(x)= x2 to the left h units (if h is negative) or to the right h units (if h is positive).
Horizontal Translations
For this equation, h = 1. The graph has the same shape as the graph of f(x)= x2, but every point has been moved to the right 1 unit.
The graph of f(x)=x2 has been horizontally translated to the right 1 unit to give the graph of f(x)= (x-1)2.
Your Turn Problem #3
Horizontally translate every point on the graph of f(x)= x2 to the left 2 units (since h=-2).
f(x)=[x-(-2)]2 x y 5 4 3 2 1
f(x)=x2
f(x)=(x+2)2
Answer:

7. There are different methods for graphing parabolas
There are different methods for graphing parabolas. Different methods for finding the vertex and different methods for finding points on the graph. What has been shown in this lesson, we can obtain the desired parabola using a vertical or horizontal translation. If the value of ‘a’ is negative, we would get a reflection, the parabola would open downward. If the value of ‘a’ is other than 1, it would change the shape of the parabola.
We will focus on another method will be more straightforward.
1. Find the vertex, which is (h,k).
2. Make a table and pick two values to the right of h, and two values to the left of h. Find the corresponding values for y= f(x).
3. Plot the points and sketch of graph of the parabola. The parabola must be symmetric.
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8. 1. Find the vertex. (h =opposite of the # inside parentheses, k = # outside parentheses.
(-1,-3)
2. Make a table. Use two numbers to the right of -1 and two numbers to the left of -1.
3. Plot and sketch.
x y 1 -2 -3 -2 1
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9. Your Turn Problem #4
x axis
y axis
Vertex (2,-1)
x y
3 0
4 3
1 0
0 3

10. x y Example 5. Solution: 1. Find the vertex. (0,1)
2. Make a table. Use two numbers to the right of 0 and
two numbers to the left of 0.
x y
1 3
2 9
-1 3
-2 9
x axis
y axis
3. Plot and sketch.
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11. Your Turn Problem #5 Vertex (1,0) x y 2 -2 3 -8 0 -2 -1 -8 y axis
x axis
y axis
Vertex (1,0)
x y
2 -2
3 -8
0 -2
-1 -8

12. If the parabola is in the form, f(x) = ax2 + bx + c, use completing the form to obtain the desired form of a parabola. An example will be shown as the procedure is given.
1. Factor the ‘a’ from the first two terms. Leave ‘c’ alone and leave a space for a number to be added later. We are creating a perfect square trinomial.
2. Take the half of the middle term and square it. Add that number inside the parentheses and add it’s opposite to the constant outside the parentheses.
Note: the number inside the parentheses is always the same number that gets squared and the same sign as the sign of the middle term.
3. Rewrite trinomial as a binomial squared and add the constants.

13. 1. Factor the ‘a’ from the first two terms
1. Factor the ‘a’ from the first two terms. Leave ‘c’ alone and leave a space for a number to be added later.
2. Take the half of the middle term and square it. Add that number inside the parentheses and add it’s opposite to the constant outside the parentheses. Because of the 2 outside the parentheses, the value being added is actually 8. Therefore a -8 will be added to the constant.
3. Rewrite trinomial as a binomial squared and add the constants.
Your Turn Problem #6

14. 1. Factor the ‘a’ from the first two terms
1. Factor the ‘a’ from the first two terms. Leave ‘c’ alone and leave a space for a number to be added later.
2. Take the half of the middle term and square it. Add that number inside the parentheses and add its opposite to the constant outside the parentheses. Because of the –5 outside the parentheses, the value being added is actually Therefore a +20 will be added to the constant outside the parentheses.
3. Rewrite trinomial as a binomial squared and add the constants.
Your Turn Problem #7

15. 1. There is no ‘a’ to factor from the first. two terms
1. There is no ‘a’ to factor from the first two terms. Leave ‘c’ alone and leave a space for a number to be added later.
2. Take the half of the middle term and square it. Add that number inside the parentheses and add it’s opposite to the constant outside the parentheses.
3. Rewrite trinomial as a binomial squared and add the constants.
Your Turn Problem #8

16. 1. Write in standard form.
2. Find the vertex.
(3,1)
3. Make a table. Use two numbers to the right of 3 and two numbers to the left of 3.
x y
4 -1
5 -7
2 -1
1 -7
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17. Your Turn Problem #9 Vertex (-2,-5) x y -1 -2 0 7 -3 -2 -4 7 The End
-1 -2
0 7
-3 -2
-4 7
The End
B.R.
1-5-07