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1. QUADRATIC FUNCTION

3. THE ZEROS OF QUADRATIC FUNCTIONS
Any value of x that makes the function f(x) equal to zero is called a zero
of the given function and consequently, a root of its corresponding
equation.
The zeros of the quadratic function y = 𝑎𝑥2 + bx + c are the roots of the
equation 𝑎𝑥2
+ bx + c = 0.
1 2 3 4
5
3
2
-1
-2
-3
-4
-5 -4 -3 -2
-1
-5
-6
Quadratic Function y = 𝑥2
− x −
6
(-2 , 0) (3 , 0)
3 and -2 are the zeros of the
function since these are the
values of x when y is equals 0.
zeros of the
function

4. THE ZEROS OF QUADRATIC FUNCTIONS
FACTORING:
𝑥2
+ bx + c
(x+_)(x+_)
COMPLETING THE
SQUARE:
𝑥2 + bx = −c
(𝑎𝑥2 + bx + _) = −c + _
Quadratic formula: 𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
Methods for finding the roots
of quadratic equations:

5. OBTAINING THE ZEROS OF A QUADRATIC
FUNCTION
Problem 1:
Find the zeros of the quadratic function y = 2𝑥2
+ 7𝑥 −
4
Solution:
y = 2𝑥2
+ 7𝑥 − 4
Thus,
2𝑥2 + 7𝑥 − 4 = 0
(2x-1)(x+4) = 0
2x-1 = 0 x+4 = 0
x =
1
2
x = -4
In order to find the
roots,
we will let y = 0
2x = 1
2 2

6. OBTAINING THE ZEROS OF A Q.FUNCTION
Thus, The zeros of f(x) are
1
2
and -4. These are the x-intercepts of the graph of
f(x) and are equivalent to the points (
1
2
, 0) and (-4 , 0). The graph of f(x)
is shown below:
(
1
2
, 0)
v(
−7
4
,
−81
8
)
(-4 , 0)
f(x) = 2𝑥2 + 7𝑥 − 4
v(
−𝑏
2𝑎
,
4𝑎𝑐 −𝑏2
4𝑎
)
v(
−7
4
,
−81
8
)
v(
−7
4
,
−32 −49
4(2)
)

7. OBTAINING THE ZEROS OF A QUADRATIC
FUNCTION
Problem 2:
Find the zeros of the quadratic function y = 4𝑥2
+ 3𝑥 − 1
Solution:
if we let y = 0 and solve the resulting equation 4𝑥2 + 3𝑥 − 1 = 0
by factoring, we get the following:
4𝑥2 + 3𝑥 − 1 = 0
(4x – 1)(x + 1) = 0
4x – 1 = 0 x + 1 = 0
x =
1
4
x = -1
4x = 1
4 4

8. OBTAINING THE ZEROS OF A QUADRATIC
FUNCTION
Thus, The zeros of f(x) are
1
4
and -1. These are the x-intercepts of the graph of
f(x) and are equivalent to the points (
1
4
, 0) and (-1 , 0). The graph of f(x) is
shown below:
(
1
4
, 0)
(-1 , 0)
f(x) = 4𝑥2 + 3𝑥 − 1
v(
−𝑏
2𝑎
,
4𝑎𝑐 −𝑏2
4𝑎
)
v(
−3
2(4)
,
4 4 −1 −(3)2
4(4)
)
v(
−3
8
,
−25
16
)
v(
−3
8
,
−25
16
)

9. OBTAINING THE ZEROS OF A QUADRATIC
FUNCTION
Problem 3:
find the zeros of a quadratic function f(x) = 𝑥2 + 𝑥 − 12.
Solution:
set y = 0.
thus, 0 = 𝑥2 + 𝑥 − 12 where a = 1, b = 1, c = -12.
𝑥 =
−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 =
−(1) ± (1)2−4(1)(−12)
2(1)
𝑥 =
−1 ± 1 + 48
2
Substitute a, b and c:
Simplify:
𝑥 =
−1 ± 49
2

10. OBTAINING THE ZEROS OF A QUADRATIC
FUNCTION
𝑥 =
−1 ± 49
2
𝑥 =
−1 + 7
2
𝑥 =
−1 − 7
2
𝑥 =
6
2
𝑥 =
−8
2
𝑥 = 3 𝑥 = −4
Therefore, the zeros of f(x) = 𝑥2 + 𝑥 − 12 are 3 and -4.

12. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Problem 4:
write a quadratic function in the form f(x) = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
when its zeros are
1
2
and -3.
Solution:
the zeros of a function are also the roots of its
corresponding equation. Thus,
=
1
2
2x = 1
2x – 1 = 0
x = -3
x + 3 = 0
x
1

13. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
2x – 1 = 0 x + 3 = 0
(2x – 1)(x + 3) = 0
2𝑥2 + 5𝑥 − 3 = 0
So, the equation with the given root is:
Thus, the function of the given zeros is f(x) = 2𝑥2 + 5𝑥 − 3.

14. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Problem 5:
Determine the equation of the parabola whose vertex is
located at (2 , -1) and whose y-intercept is equal to -3.
Solution:
since the vertex (h , K) is given, use the equation f(x) = 𝑎(𝑥 − ℎ)2 + 𝐾.
Substitute: v(2 , -1)
f(x) = 𝑎(𝑥 − 2)2 − 1
A y-intercept of -3 means that the graph crosses the y-axis at the
point (0 , -3). Substitute x = 0 and f(x) = -3 in the equation.

15. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
f(x) = 𝑎(𝑥 − 2)2 − 1
-3 = 𝑎(0 − 2)2
− 1
-3 = 𝑎(2)2 − 1
-3 = 4𝑎 − 1
-4a = 3 − 1
-4a = 2
−4 − 4
a = −
1
2
The equation of the given
parabola is f(x) = −
1
2
(𝑥 − 2)2 −
1.
Substitute:
f(x) = −
1
2
(𝑥 − 2)2 − 1
Substitute the value of a from the
equation:

16. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Problem 6:
find the equation of the quadratic function determined
from the graph below:
0 1 2 3 4 5
-1
-2
-3
-4
2
1
-2 -1
(-1 , 0) (5 , 0)
(2 , -3)

17. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Solution:
the vertex of the graph of the function is (2 , -3). The graph
passes through the point (5 , 0). By replacing x and y with 5 and 0,
respectively, and h and K with 2 and -3, respectively, we have:
Y = 𝑎(𝑥 − ℎ)2 − 𝐾
0 = 𝑎(5 − 2)2 − 3
0 = 𝑎(3)2 − 3
0 =9𝑎 − 3
-9a = -3
a =
1
3
−9 − 9
Thus, the quadratic function
is y =
1
3
(𝑥 − 2)2 − 3.
Substitute the values of
x and y:

18. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Problem 7:
find a quadratic function whose zeros are -1 and
4.
Solution:
if the zeros are 𝑟1 = -1 and 𝑟2 = 4, then:
f(x) = (x - 𝑟1)(x - 𝑟2)
f(x) = [x – (-1)] (x - 4) Thus, the function is
f(x) = 𝑥2 − 3𝑥 − 4
f(x) = (x + 1) (x - 4)

19. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
But the equation f(x) = 𝑥2 − 3𝑥 − 4 is not
unique since there are other quadratic functions
whose zeros are -1 and 4 such as f(x) = 2𝑥2 − 6𝑥 −
8 and f(x) = 3𝑥2 − 9𝑥 − 12 and many more.
These equations are obtained by multiplying the
right hand side of the equation by a nonzero
constant.
Thus, the answer is f(x) = a(𝑥2 − 3𝑥 − 4), where a
is any nonzero constant.

20. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Problem 8:
find the equation of a quadratic function whose zeros are
3 ± 2
3
.
Solution:
A quadratic equation with irrational roots cannot be
written as a product of linear factors with irrational coefficients.
In this case, we can use another method. Since the zeros are
3 ± 2
3
then:
=
3 ± 2
31
x
3x = 3 ± 2
3x - 3 = ± 2

21. FINDING THE EQUATION OF A QUADRATIC
FUNCTION
Square both sides of the equation and simplify:
3x - 3 = 2( )2 ( )2
Therefore we get:
3𝑥2 − 18𝑥 + 9 = 2
3𝑥2
− 18𝑥 + 7 = 0
Thus, the equation of quadratic function
is f(x) = 3𝑥2 − 18𝑥 + 9.