Identifying Quadratic
Functions
Lesson 8-1
What does a quadratic equation (or function)
look like?
• When we are given an equation (or function), it needs to be of the
form 𝒙𝟐 + 𝑏𝑥 + 𝑐 or 𝒂𝒙𝟐 + 𝑏𝑥 + 𝑐.
• IT MUST HAVE THE 𝑿𝟐 TERM TO BE QUADRATIC!! THAT IS WHY IT IS
IN RED!
• IF THERE IS NO 𝑿𝟐 TERM, 𝒙𝟐 = 𝟎 𝒐𝒓 𝒂𝒙𝟐 = 𝟎.
• In a quadratic equation, as long as that 𝒙𝟐 term is there, the
equation (or function) is quadratic.
Example 1: Tell whether the equation (or
function) is quadratic.
• y = -3x + 20
• When we are given an equation (or function), it needs to be of the
form 𝒙𝟐 + 𝑏𝑥 + 𝑐 or 𝒂𝒙𝟐 + 𝑏𝑥 + 𝑐.
• Is there an 𝒙𝟐 term?
• NO THERE IS NOT.
• SO THIS IS NOT A QUADRATIC EQUATION/FUNCTION.
• A QUADRATIC EQUATION/FUNCTION MUST HAVE AN 𝑿𝟐 TERM.
Example 2: Tell whether the equation (or
function) is quadratic continued.
• 𝑦 + 3𝑥 2 = −4
• Write it in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
• 𝑦 + 3𝑥 2 = −4
Get y by itself.
• −3𝑥 2 −3𝑥 2
Subtract 𝟑𝒙𝟐 from both sides.
• Our new equation is 𝑦 = −3𝑥 2 − 4.
• Is there an 𝑥 2 term?
• YES THERE IS AN 𝑋 2 TERM!!!
• 𝑦 = −3𝑥 2 − 4 is a quadratic function!
Example 3: Tell whether the equation (or
function) is quadratic continued.
• 𝑦 + 𝑥 = 2𝑥 2
• Write it in the form a𝑥 2 + 𝑏𝑥 + 𝑐.
• 𝑦 + 𝑥 = 2𝑥 2
Get y by itself.
• −𝑥 −𝑥
Subtract x from both sides.
• Our new equation is 𝑦 = 2𝑥 2 − 𝑥.
• Is there an 𝑥 2 term?
• YES THERE IS!!!
• 𝑦 = 2𝑥 2 − 𝑥 is a quadratic function!
Your Turn: Tell whether the function is
quadratic.
• 𝑦 + 6𝑥 = −14
2𝑥 2 + 𝑦 = 3𝑥 − 1
• Remember to get y by itself
• See if there is an 𝑥 2 term.
Example 4: Graph using a table of values
1) Fill the table of values to complete the table.
2) Plot the points on a graph and draw the parabola.
The equation is 𝑦 = 2𝑥 2 .
Use the given x-values to complete
the table.
x
𝒚 = 𝟐𝒙𝟐
-2
8
-1
2
0
0
1
2
2
8
Example 5: Graph using a table of values
1) Fill the table of values to complete the table.
2) Plot the points on a graph and draw the parabola.
The equation is 𝑦 = −2𝑥 2 .
Use the given x-values to complete
the table.
x
𝒚 = −𝟐𝒙𝟐
-2
-8
-1
-2
0
0
1
-2
2
-8
Your turn: Graph using a table of values.
• 𝑦 = 𝑥2 + 2
• Use the given x-values.
x
𝒚 = 𝒙𝟐 + 𝟐
-2
6
-1
3
0
2
1
3
2
6
Your turn: Graph using a table of values.
• 𝑦 = −3𝑥 2 + 1
• Use the given x-values.
x
𝒚 = −𝟑𝒙𝟐 + 𝟏
-2
-11
-1
-2
0
1
1
-2
2
-11
Identifying the direction of the parabola when
𝑎 > 0.
• Given the function a𝑥 2 + 𝑏𝑥 + 𝑐, if 𝑎 > 0, the parabola opens
upward, and the y-value of the vertex is the minimum value of the
function.
1) The vertex is the lowest point on
the parabola when 𝑎 > 0. The point
in purple is the vertex.
2) The coordinates of the vertex are
(-5,-7).
3) If the parabola opens upward, the
vertex is the minimum value.
4) The minimum value is the ycoordinate of the vertex.
5) So the minimum value would be -7.
Identifying the direction of the parabola when
𝑎 < 0.
• Given the function a𝑥 2 + 𝑏𝑥 + 𝑐, if 𝑎 < 0, the parabola opens
downward, and the y-value of the vertex is the maximum value of the
function.
1) The vertex is the highest point on
the parabola when 𝑎 < 0. The point
in purple is the vertex.
2) The coordinates of the vertex are
(4,0).
3) If the parabola opens downward,
the vertex is the maximum value.
4) The maximum value is the ycoordinate of the vertex.
5) So the maximum value would be 0.
Example 6: Tell whether the graph of each
quadratic function opens upward or downward.
•
•
•
•
•
•
•
•
•
•
𝑦 = 4𝑥 2
𝑎=4
Is 4 > 0 or is 4 < 0?
Yes it is. So the parabola opens upward.
2𝑥 2 + 𝑦 = 5
We must get y by itself first.
−2𝑥 2
−2𝑥 2
Subtract 2𝑥 2 from both sides.
Our new function is 𝑦 = −2𝑥 2 + 5.
𝑎 = −2
Is −2 > 0 or is −2 < 0?
−2 < 0, so the parabola opens downward.
Example 7: Identify the vertex of the parabola.
Then state the maximum or minimum value.
1) What is the vertex?
2) Is it a maximum or minimum?
3) What is the maximum value or the
minimum value?
Example 8: Identify the vertex of the parabola.
Then state the maximum or minimum value.
1) What is the vertex?
2) Is it a maximum or minimum?
3) What is the maximum value or the
minimum value?
Domain and range of a quadratic function.
Domain: All real numbers
Range: 𝑦 ≥ 5
1) When the parabola opens upward,
the range begins with the minimum
value.
2) The vertex is at the point (3,5).
3) The minimum value is the ycoordinate of the vertex.
4) The minimum value in this parabola
is 5.
5) When the parabola opens upward,
the range is the minimum value and
everything above it.
6) The range can be said to be 𝑦 ≥ 5.
7) Unless a specific set of numbers is
given, assume the domain is “all
real numbers.”
Domain and range of a quadratic function.
Domain: All real numbers
Range: 𝑦 ≤ 2
1) When the parabola opens
downward, the range begins with
the maximum value.
2) The vertex is at the point (-6,2).
3) The maximum value is the ycoordinate of the vertex.
4) The maximum value is 2.
5) When the parabola opens
downward, the range is the
maximum value and everything
below it.
6) The range can be said to be 𝑦 ≤ 2.
7) Unless a specific set of numbers is
given, assume the domain is “all
real numbers.”
Recap…
• The 𝑥 2 term must be there for the equation (or function) to be quadratic.
• When graphing, a table needs to be made. Use specific values of ‘x’ to plug
into the equation. Then plot the points and draw the parabola.
• When 𝑎 > 0, the parabola opens upward and the y-coordinate of the
vertex is the minimum value.
• When 𝑎 < 0, the parabola opens downward and the y-coordinate of the
vertex is the maximum value.
• Look at the number in front of the 𝑥 2 term to see if it opens upward or
downward.
• Assume the domain is “all real numbers” unless otherwise stated.
Recap…
• When the parabola opens upward, use 𝑦 ≥ 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 for the
range.
• When the parabola opens downward, use 𝑦 ≤ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 for
the range.
Homework
• Page 527 (26, 28, 30 – 38 all)
Characteristics of Quadratic
Functions
Lesson 8-2
Example 1: Finding zeros of quadratic
functions from a graph.
• A zero of a function is an x-value that makes the function equal to 0.
• In other words, the zero of a function is the x-intercept – where the
graph crosses the x-axis – of the function.
𝑦 = 𝑥 2 − 2𝑥 − 3
1. To see what the zeros are in the graph,
see where the parabola crosses the xaxis..
2. The parabola crosses the x-axis at the
points 3 and -1.
3. So, the zeros of the function are 𝑥 =
3 𝑜𝑟 𝑥 = −1.
Example 2: Finding zeros of quadratic
functions from a graph.
• A zero of a function is an x-value that makes the function equal to 0.
• In other words, the zero of a function is the x-intercept – where the
graph crosses the x-axis – of the function.
𝑦 = 𝑥 2 + 8𝑥 + 16
1. To see what the zeros are in the graph,
see where the parabola crosses the xaxis..
2. The parabola crosses the x-axis at the
point -4.
3. So, the zero of the function is 𝑥 = −4
Example 3: Finding zeros of quadratic
functions from a graph.
1. To see what the zeros are in the graph,
see where the parabola crosses the xaxis..
2. The parabola crosses the x-axis at the
points 4 and -2.
3. So, the zeros of the function are 𝑥 =
4 𝑜𝑟 𝑥 = −2.
Example 4: Finding zeros of quadratic
functions from a graph.
𝑦 = −4𝑥 2 −2
1. To see what the zeros are in the graph,
see where the parabola crosses the xaxis..
2. The parabola does NOT cross the x-axis!!!
3. So, there are no zeros of the function.
Finding the axis of symmetry.
• The axis of symmetry is a vertical line that divides the parabola into
two symmetrical sides.
• The axis of symmetry passes through the vertex. We are going to use
the zeros to find the axis of symmetry.
Example 5: Find the axis of symmetry of the
parabola by using zeros.
1. This parabola has ONE zero, so we
use the x-coordinate of the vertex.
2. What is the vertex?
3. The vertex is at the point (-1,0).
4. The x-coordinate of the vertex is -1.
5. The axis of symmetry is 𝑥 = −1.
Example 6: Find the axis of symmetry of the
parabola by using zeros.
1. This parabola has TWO zeros, so we
use the average of the two zeros to
find the axis of symmetry.
2. What are the two zeros?
3. The two zeros are 𝑥 = 1 𝑜𝑟 𝑥 = 4.
4. We now compute the average of
these two zeros.
5. The average is the sum of the two
zeros divided by two.
1+4
1. The average is
2
=
5
.
2
6. The axis of symmetry is at 𝑥 =
5
.
2
Formula for computing the axis of symmetry.
• You can use this formula if there are no zeros or the zeros are hard to
identify from a graph. This formula works for all quadratic functions.
Example 7: Find the axis of symmetry by using
the formula.
• 𝑦 = 1𝑥 2 + 3𝑥 + 4
• Find the values of a and b.
• 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 3
• Use the new formula we are given.
•𝑥 =
•𝑥 =
𝑏
−
2𝑎
3
−
2 1
=
3
−
2
• The axis of symmetry is 𝒙 =
𝟑
− .
𝟐
Example 8: Find the axis of symmetry by using
the formula.
•
•
•
•
•
𝑦 = 3𝑥 2 − 18𝑥 + 1
Find the values of a and b.
𝑎 = 3 𝑎𝑛𝑑 𝑏 = −18
DO NOT FORGET THE NEGATIVE SINCE IT IS SUBTRACTION!!!
Use the new formula we are given.
•𝑥=
𝑏
−
2𝑎
−18
−
2 3
−18
−
6
•𝑥=
=
=3
• Two negatives make a positive!!
• The axis of symmetry is 𝒙 = 𝟑.
Practice (Still part of notes).
• Page 535 (3-6, 8, 9, 12)
Finding the vertex of a parabola
• Since we know how to find the axis of symmetry, we can now find the
vertex of a parabola using the following steps.
1. Find the axis of symmetry of the parabola (x =).
2. Plug in the x-value into the equation to find the y-value.
3. Write the vertex as an ordered pair.
Example 9: Find the vertex of the parabola.
• 𝑦 = 5𝑥 2 − 10𝑥 + 3
• Step One: Find the axis of symmetry with the formula.
•𝑥 =
•𝑥 =
𝑏
−
2𝑎
−10
−
2 5
−10
−
10
•𝑥 =
= − −1 = 1
• TWO NEGATIVES MAKE A POSITIVE.
• The axis of symmetry is 𝑥 = 1.
Example 9 Continued: Find the vertex of the
parabola.
• 𝑦 = 5𝑥 2 − 10𝑥 + 3
• Step Two: Plug in the axis of symmetry into the above equation.
• 𝑥 = 1 (Axis of Symmetry)
• 𝑦 = 5(1)2 −10 1 + 3
• y = 5 − 10 + 3
• 𝑦 = −5 + 3
• 𝑦 = −2
• The axis of symmetry is 𝑥 = 1 and the y-value is 𝑦 = −2.
Example 9 Continued: Find the vertex of the
parabola.
• 𝑦 = 5𝑥 2 − 10𝑥 + 3
• Step Three: Use the two coordinates to make the ordered pair for the
vertex.
• 𝑥 = 1 (Axis of Symmetry)
• 𝑦 = −2
• The vertex is at the point (1, −2).
Example 10: Find the vertex of the parabola.
• 𝑦 = −𝑥 2 + 6𝑥 + 1
• Step One: Find the axis of symmetry.
•𝑥 =
•𝑥 =
𝑏
−
2𝑎
6
−
2 −1
•𝑥 =3
=
6
−
−2
=3
Example 10 Continued: Find the vertex of the
parabola.
• 𝑦 = −𝑥 2 + 6𝑥 + 1
• Step Two: Find the y-value.
•𝑥=3
(Axis of Symmetry)
• 𝑦 = −32 + 6 3 + 1
• 𝑦 = −9 + 18 + 1
•𝑦 =9+1
• 𝑦 = 10
• Step Three: Write the ordered pair for the vertex.
• (3,10) –FINISHED WITH THIS PROBLEM
Homework
• Page 536 (19-23, 25, 26, 29-31)
Transforming Quadratic
Functions
Lesson 8-4
The basics…
• We need to know the “parent” function. For quadratic functions, the
parent function is 𝑓 𝑥 = 𝑥 2 .
• Every other quadratic function is a transformation of the graph of
𝑓 𝑥 = 𝑥2.
• For 𝑓 𝑥 = 𝑥 2 , the axis of symmetry is 𝑥 = 0.
• The vertex is (0,0).
• The function has only one zero.
• We will be comparing:
• Coefficient of the 𝑥 2 term
• Vertical translations
𝟐
Coefficient of the 𝒙 term
• We first learned that in the function 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, if 𝑎 < 0, the
parabola opens downward and if 𝑎 > 0, the parabola opens upward.
• Now we will learn a new use of a.
𝟐
Coefficient of the 𝒙 term continued.
• The bigger the number, the skinnier the parabola.
• The smaller the number, the fatter the parabola.
Example 1: Order the functions from skinniest
to fattest.
• 𝑓 𝑥 =
−2𝑥 2 , 𝑔
𝑥 =
1 2
𝑥 ,ℎ
3
𝑥 = 4𝑥 2
• Find 𝑎 for each function.
• −2 = 2,
•
•
•
•
1
3
=
1
,
3
4 =4
Which function has the coefficient that is the biggest?
ℎ 𝑥 = 4𝑥 2
The next?
𝑓 𝑥 = −2𝑥 2
• From skinniest to fattest: ℎ 𝑥 =
4𝑥 2 , 𝑓
𝑥 =
−2𝑥 2 , 𝑔
𝑥 =
1 2
𝑥
3
• We would do the same thing if we were going from fattest to skinniest.
Vertical Translations
Example 2: Compare the following function to
2
the parent function 𝑓 𝑥 = 𝑥 .
• 𝑓 𝑥 = 𝑥2 + 5
• The function 𝑓 𝑥 = 𝑥 2 + 5 is translated 5 units up.
• 𝑔 𝑥 = −2𝑥 2 + 8
• The function 𝑔 𝑥 = −2𝑥 2 + 8 is translated 8 units up.
•ℎ 𝑥 =
3 2
𝑥
4
− 12
• The function ℎ 𝑥 =
3 2
𝑥
4
− 12 is translated 12 units down.
• WHEN ADDING, THE FUNCTION IS TRANSLATED UP. WHEN
SUBTRACTING, THE FUNCTION IS TRANSLATED DOWN.
Recap…
• To see if the parabola gets fatter, see if 𝑎 < 1.
• To see if the parabola gets skinnier, see if 𝑎 > 1.
• To order the functions from least to greatest or visa versa, compare
𝑎 from all of the functions.
• To see if the graph is translated up or down, check to see if it is adding
or subtracting.
Homework
• Page 549 (10-17)
• Page 553 (1-16, 21-24) (I know it says quiz at the top, but this is your
study guide for the test!)
Solving Quadratic Equations
by Factoring
Lesson 8-6
Remember Factoring?
• 𝑥 2 + 4𝑥 − 12
• We first set up our factoring as (𝑥
)(𝑥
).
• We found out which numbers multiplied to get −12 and add to get 4.
• Then we filled in the blanks.
• What two numbers multiply to give us −12?
• 6 ∗ −2 will add to get 4.
• Now we fill in the blanks: (𝑥 + 6)(𝑥 − 2).
• You will do two of these on your own.
• 𝑥 2 − 6𝑥 + 8
and
𝑥 2 + 9𝑥 + 20
Example 1:Solving quadratics by factoring.
• 𝑥 2 − 8𝑥 − 9
• We first set up our factoring as (𝑥
)(𝑥
).
• We find out which numbers multiply to get −9 and add to get −8.
• Then we fill in the blanks.
• What two numbers multiply to give us −9 and add to get −8?
• −9 ∗ 1 will add to give us −8.
• Now we fill in the blanks: (𝑥 − 9)(𝑥 + 1).
• Now set this equal to 0. 𝑥 − 9 𝑥 + 1 = 0.
Example 1: Solving quadratics by factoring
continued.
• 𝑥−9 𝑥+1 =0
• Now we just set 𝑥 − 9 = 0 or 𝑥 + 1 = 0
• This is called the Zero Product Property.
• If the product of two quantities is equal to zero, at least one of the
quantities equals zero.
• Solve for x: 𝑥 − 9 = 0 or 𝑥 + 1 = 0.
•
+9 +9
−1 −1
• The solutions to 𝑥 2 − 8𝑥 − 9 are 𝑥 = 9 or 𝑥 = −1.
Example 2:
• 𝑥 − 3 𝑥 + 7 = 0 THERE IS NO NEED TO FACTOR HERE. IT IS
ALREADY IN FACTORED FORM.
• We just set 𝑥 − 3 = 0 or 𝑥 + 7 = 0
• Solve for x: 𝑥 − 3 = 0 or 𝑥 + 7 = 0.
•
+3 +3
−7 −7
• The solutions to 𝑥 − 3 𝑥 + 7 are 𝑥 = 3 or 𝑥 = −7.
Example 3:
• 𝑥 𝑥 − 5 = 0 THERE IS NO NEED TO FACTOR HERE. IT IS ALREADY
IN FACTORED FORM.
• We just set 𝑥 = 0 and 𝑥 − 5 = 0
• Solve for x: 𝑥 = 0 [X IS ALREADY BY ITSELF HERE] or 𝑥 − 5 = 0.
•
+5 +5
• The solutions to 𝑥 𝑥 − 5 are 𝑥 = 0 or 𝑥 = 5.
Example 4:
• −2𝑥 2 = 18 − 12𝑥 WE MUST GET THIS IN STANDARD FORM.
• Since we are factoring, we need the 𝑥 2 term to be positive.
• So we have to move the −2𝑥 2 to the other side.
• −2𝑥 2 = 18 − 12𝑥
• +2𝑥 2
+2𝑥 2
• Our new equation is 2𝑥 2 − 12𝑥 + 18. What is the GCF of 2, 12, and
18?
• GCF = 2. Divide every number in the equation by 2.
• 2(𝑥 2 − 6𝑥 + 9) Now factor the trinomial.
Example 4 Continued:
• 2(𝑥 2 − 6𝑥 + 9) DO NOT FORGET ABOUT THAT 2 OUT FRONT
• First start with 2(𝑥
)(𝑥
).
• Now find the factors of 9 that add up to −6.
• −3 ∗ −3 multiply to give you 9 and add to give you −6.
• Fill in the blanks: 2(𝑥 − 3)(𝑥 − 3).
• Set them equal to zero: 𝑥 − 3 = 0 or 𝑥 − 3 = 0.
• 2 ≠ 0, so solve for x: 𝑥 − 3 = 0
•
+3 +3
• The only solution is 𝑥 = 3.
Example 5:
• 2𝑥 + 1 3𝑥 − 1 = 0
• Set both equal to 0.
• Solve for x: 2𝑥 + 1 = 0
• Solve for x: 3𝑥 − 1 = 0
•𝑥=
1
−
2
or 𝑥 =
1
3
Example 6:
•
•
•
•
•
•
•
•
𝑥
4𝑥 2 − 9𝑥 + 2 = 0
Use the box for this.
4𝑥
4𝑥 2
Multiply the first and last term.
4∗2=8
−1 −1𝑥
Find two factors of 8 that add to get −9.
−1 ∗ −8: Put these two in the box along with an x.
Our factors are 4𝑥 − 1 𝑥 − 2 = 0.
So, 4𝑥 − 1 = 0 or 𝑥 − 2 = 0. Solve for x.
1
4
• 𝑥 = or 𝑥 = 2
−2
−8𝑥
2
Homework
• Page 565 (2-18)
Solving Quadratic Equations
by Using Square Roots
Lesson 8-7
Square Roots
• Every positive real number has two square roots.
Positive
Square root of 9
Negative
Square root of 9
Positive and negative
Square roots of 9
When writing square roots, we will always use the ± sign in front of
the number. Just like with the three above.
Example 1:
• 𝑥 2 = 16.
• To get 𝑥 by itself, we must take the square root of both sides.
• 𝑥 2 = 16
• What is the square root of 𝑥 2 ? What is the square root of 16?
• Always use the ± sign.
• 𝑥 = ±4
• This means that the two solutions are 𝑥 = 4 and 𝑥 = −4.
Example 2:
• 𝑥 2 = −81.
• To get 𝑥 by itself, we must take the square root of both sides.
• 𝑥 2 = −81
• What is the square root of 𝑥 2 ? What is the square root of −81?
• Always use the ± sign.
• Is there any number times itself that will give you −81?
• NO THERE IS NOT!! THIS QUADRATIC EQUATION HAS NO REAL
SOLUTIONS.
Example 3:
• 4𝑥 2 − 25 = 0
• We must get 𝑥 by itself.
• 4𝑥 2 − 25 = 0
•
+25 + 25
• Our new equation is 4𝑥 2 = 25
• Get the 𝑥 2 by itself.
• 𝑥2 =
25
4
• Now take the square root of both sides.
•
25
4
𝑥2 =
• 𝑥=±
5
2
Example 4:
• (𝑥 − 7)2 = 9.
• To get 𝑥 by itself, we must take the square root of both sides.
• (𝑥 − 7)2 = 9
• What is the square root of (𝑥 − 7)2 ? What is the square root of 9?
• The square root and the square on the left side cancel out.
• So our new equation is 𝑥 − 7 = ±3. Since we have the positive and
negative three, we will have two solutions.
• Always use the ± sign.
• 𝑥 − 7 = 3 or 𝑥 − 7 = −3. Solve for 𝑥.
• The solutions are 𝑥 = 10 or 𝑥 = 4
Example 4:
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100 − 5𝑥 2 = 0.
Solve for 𝑥.
100 − 5𝑥 2 = 0
−100
− 100
−5𝑥 2 = −100
Divide both sides by −5. TWO NEGATIVES MAKE A POSITIVE.
𝑥 2 = 20
• 𝑥 2 = 20
• The solutions are 𝑥 = ± 20.
Homework
• Page 571 (18-34 even)