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Objective: To find the zeros of a quadratic
function and solve a quadratic equation
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Quadratic Equation:
◦ A second-degree equation in one variable.
◦ Has the form ax2 + bx + c = 0
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Quadratic Function:
◦ A second-degree function
◦ Has the form f(x) = ax2 + bx + c
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Zero of a Function:
◦ The point or points where the graph of the function
crosses the x-axis
◦ Can be found by determining where f(x) = 0
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Zero-Product Property:
◦ If a multiplication problem has a product of zero,
the one of the factors MUST equal zero
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Solution of a Quadratic Equation:
◦ The value(s) of x which satisfy the equation
◦ The solutions of any quadratic equation are the
same as the zeros of the related quadratic function
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When finding the zeros of a function or the
solutions to an equation, the number of zeros
or solutions will be equal to the degree of the
function/equation.
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Graphing
Factoring
Using square roots
Completing the square
Using the Quadratic Formula
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Use the graph below to determine the zeros
of the function f(x) = x2 – 3x - 10
y
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The zeros of the
function are -2 and 5
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x
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Determine the zeros f(x) = x2 – 3x – 10 by
factoring
◦
◦
◦
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Set the function equal to zero, then factor
x2 – 3x – 10 = 0
(x – 5)(x + 2) = 0
Use the zero-product property to determine the
zeros
x–5=0
x=5
or
x+2=0
x = -2
The zeros of the function are
-2 and 5
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Determine the zeros of f(x) = 2x2 + 3x – 9 by
factoring
2x2 + 3x – 9 = 0
(2x - 3)(x + 3) = 0
2x – 3 = 0
2x = 3
3
x
2
or
x+3=0
x = -3
The zeros of the function are
-3 and 3/2
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Determine the zeros of f(x) = 9x2 – 30x + 25 by
factoring
9x2 - 30x + 25 = 0
(3x - 5)2 = 0
y
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(3x – 5)2 = 0
x
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(3x  5)  0
2
3x – 5 = 0
3x = 5
5
x
3
This function has a double
zero of 5/3
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The graph of the function touches
the x-axis at the double zero,
rather than crossing it. This is
called a point of tangency.
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Determine the zeros of f(x) = 16x2 – 49 by
factoring
16x2 – 49 = 0
(4x + 7)(4x – 7) = 0
4x + 7 = 0
4x = -7
x
7
4
or
4x - 7 = 0
4x = 7
7
x
4
The zeros of the function are
-7/4 and 7/4
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Determine the zeros of f(x) = 16x2 – 49 by
using square roots
◦ Because this function has no first-degree term, it
can be solved by using square roots
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16x2 – 49 = 0
16x2  49
49
x 
16
2
49
x 
16
2
x
7
4
The zeros of the function are
-7/4 and 7/4
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Determine the zeros of f(x) = x2 + 4x – 2
◦ This function cannot be factored
◦ This function cannot be solved by using square
roots
◦ Graphing the function will show that the zeros are
non-terminating, non-repeating decimals
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The exact values of the zeros of this function
can be found by using a method called
Completing the Square
Completing the square is a process which
turns the function into a perfect square
trinomial
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To complete the square, use the following
steps
◦ Set the function equal to zero
◦ Move the constant to the other side of the equation
◦ If the leading coefficient is not 1, divide each term
by the leading coefficient
◦ Determine the number needed to “complete the
square” by taking ½ of the first-degree term’s
coefficient, squaring it, and adding it to both sides
of the equation
◦ Factor the resulting perfect square trinomial
◦ Determine the zeros by using square roots
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Determine the zeros of f(x) = x2 + 4x – 2
x2 + 4x – 2 = 0
Take ½ of 4 which is 2, square it
2
+
4
x + 4x + 4 = 2
to get 4, then add 4 to both sides
(x + 2)2 = 6
 x  2
2
 6
x 2   6
x  2  6
The zeros of this function
are 2  6 and 2  6
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Determine the zeros of f(x) = x2 - 6x + 4
x2 - 6x + 4 = 0
Take ½ of -6 which is -3, square it
2
+
9
x - 6x + 9= -4
to get 9, then add 9 to both sides
(x - 3)2 = 5
 x  3
2
 5
x 3  5
x  3 5
The zeros of this function
are 3  5 and 3  5
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Determine the zeros of f(x) = 2x2 - 5x + 1
As the leading coefficient is not 1,
2x2 - 5x + 1 = 0
divide both sides by 2
2
2x – 5x
= -1
Take ½ of -5/2 which is -5/4,
5
1  25
25
2
square it to get 25/16, then add
x  x

16
16
25/16 to both sides
2
2
2
5  17

x   
5
17
4  16
The zeros of

x 
this function
4
4
2
5
17
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are 5  17
5
17
x
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x 
4
16

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4
4
4
5  17
and
4
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Solving a quadratic equation is very similar to
finding the zeros of a quadratic function
◦ Get the equation into standard form
◦ Solve by graphing, factoring, using square roots or
completing the square
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Solve:
3x2 – 2x = 1
3x2 – 2x – 1 = 0
(3x + 1)(x – 1) = 0
3x + 1 = 0
or
3x = -1
x = -1/3
The solution set to
x–1=0
x=1
x=1
this equation is {-1/3, 1}
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Solve:
4x2 – 6x = 0
2x(2x – 3) = 0
2x = 0
or
x=0
x=0
The solution set to
2x – 3 = 0
2x = 3
x = 3/2
this equation is {0, 3/2}
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Solve:
3(x – 2)2 – 6 = 0
3(x – 2)2 = 6
(x – 2)2 = 2
x–2=  2
x = 2 2
The solution set to this equation is { 2  2 }

The graph of a quadratic function is a
parabola and can be graphed using its zeros.
◦ Determine the zeros using any method.
◦ Plot the zeros on the graph.
◦ The vertex of the parabola is halfway between the
zeros.
 Determine the midpoint of the zeros. This is the xvalue of the vertex.
 Substitute into the function to determine the y-value
of the vertex.
◦ The y-intercept of the parabola can be found by
substituting 0 for x and solving for y.
◦ Use the points to sketch the graph of the parabola.
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Graph f(x) = x2 – 2x – 8
Factor: f(x) = (x – 4)(x + 2)
The zeros of the function
are 4 and -2
The midpoint of the zeros
is 1
Plug 1 in for x and get -9,
so the vertex is (1, -9)
Plug zero in for x and get
-8, so the y-intercept is
(0, -8)
Sketch
y
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x
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