Quadratic Function:

The quadratic equation or parabolic function has a shape of

equation as:
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
As seen the highest exponential is 2.
a, b and c are factors of the quadratic equation, and the graph
shows a parabolic shape.
If (a) the leading factor is:
1. Positive a>0: the graph is concave upward.
2. Negative a<0: the graph is concave down.

The graph shows parts of the parabolic function.

As seen, when the graph intersects the x-axis it is
called: Roots, zeros or solutions.
  Finding Roots:
Roots are the x-intercept of the graph (y=0), here are cases to make it easier to
remember:
1. 𝒂𝒙𝟐 − 𝒄 = 𝟎:
In this case just (a) the factor of 𝑥 2 and (c) is given, as an example:
Example (1): What are the zeros for 𝑦 = 2𝑥 2 − 18?
Solution:
2𝑥 2 = 18

𝑥 2 = 9 , 𝑡ℎ𝑒𝑛 √𝑥 2 = √9
𝑥 = ±3
Answer: Zeros are (−3,0) 𝑎𝑛𝑑 (3,0)

2. 𝒂𝒙𝟐 + 𝒃𝒙 = 𝟎
This case (a) and (b) factors giver, to solve use common factor:
Example (2): find the x-intercept of 𝑓(𝑥) = 2𝑥 2 + 18𝑥?
Solution:
Make y=0
2𝑥 2 + 18𝑥 = 0
Common factor is 2𝑥:
2𝑥(𝑥 + 9) = 0
Then; two terms when multiply the result is zero, either the first term is zero or the
second.
2𝑥 = 0, 𝑡ℎ𝑒𝑛 𝑥 = 0
Or 𝑥 + 9 = 0 , 𝑡ℎ𝑒𝑛 𝑥 = −9
As a points:
𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 (0,0) 𝑎𝑛𝑑 (−9,0)
 3. 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎
In this case, two main methods can be used to solve this equations:
 Factoring
 General law (quadratic formula)

Factoring:
Is to convert the equation into two multiplied brackets giving the same equation once
re-multiply.
Look at the figure below to show the factoring method.

Some equation can’t be factorized, to solve them general formula can be used.

Quadratic Formula:

This is the quadratic formula that can define the value of x for any equation, where a,b
and c given from the equation, and √𝑏 2 − 4𝑎𝑐 is called discriminant.
Discriminant:

The figure above shows the three cases that discriminant can know, it helps to know the
number of solutions and predict the graph of the equation.
After finding the discriminan0t, use its value to substitute in the quadratic formula to
find the value of x-intercepts (Roots):

Note: calculator can be used to find the roots of a quadratic function.

Hint: when the question have real roots or non-real roots use discriminant.

The next 3 examples will explain the methods of solving a quadratic equations (factoring
& quadratic formula) and will show the three cases of answers.
Example (3): find the roots of 𝑦 = 𝑥 2 − 7𝑥 + 10?
Solution:
1st method factorize:
𝑥 2 − 7𝑥 + 10 = 0
(𝑥 − 5)(𝑥 − 2) = 0
Then:
𝑥 − 5 = 0 , 𝑎𝑛𝑑 𝑥 = 5
Or
𝑥 − 2 = 0 , 𝑎𝑛𝑑 𝑥 = 2
As points (2,0) and (5,0).

2nd method quadratic formula:

𝑥 2 − 7𝑥 + 10 = 0
𝑎 = 1 , 𝑏 = −7 , 𝑐 = 10
Find the discriminant:

√𝑏 2 − 4𝑎𝑐 = √(−7)2 − 4(1)(10) = √49 − 40 = √9 = 3 > 0 , it has two real roots

because discriminant is greater than zero.

−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑋1,2 =
2𝑎
−(−7) + 3 10
𝑥1 = = = 5, (5,0)
2(1) 2
−(−7) − 3 4
𝑥2 = = = 2, (2,0)
2(1) 2
Example (4): If 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 16, what are the zeros?
Solution:
1st method factorize:
𝑥 2 − 8𝑥 + 16 = 0
(𝑥 − 4)(𝑥 − 4) = 0
𝑥 − 4 = 0, 𝑡ℎ𝑒𝑛 𝑥 = 4
(4,0)
Both x are equal to 4 that means the equation has two equal roots (one real solution).

2nd method quadratic formula:

𝑥 2 − 8𝑥 + 16
𝑎 = 1, 𝑏 = −8 𝑎𝑛𝑑 𝑐 = 16
Find the discriminant:

√𝑏 2 − 4𝑎𝑐 = √(−8)2 − 4(1)(16) = √64 − 64 = √0 = 0

The discriminant is zero, therefore the equation has one real root or two equal roots.

−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑋1,2 =
2𝑎
−(−8) ± 0 8
𝑥1,2 = = = 4, (4,0)
2(1) 2

Example (5): What are the x-intercepts for 𝑦 = 𝑥 2 + 2𝑥 + 2?

Solution:
Factorize method can’t be applied, so quadratic formula will be used:
𝑥 2 2𝑥 + 2
𝑎 = 1, 𝑏 = 2, 𝑐 = 2
Discriminant:

√𝑏 2 − 4𝑎𝑐 = √(2)2 − 4(1)(2) = √4 − 8 = √−4 < 0

The value of the discriminant is less than zero , means it has no real solutions and the
solution is imaginary (complex).
 √−4 = √−1 × √4 = ±2𝑖

Because (√−1 = 𝑖) imaginary number.

−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑋1,2 =
2𝑎
−(2) + 2𝑖
𝑥1 = = −1 + 𝑖
2(1)
−(2) − 2𝑖
𝑥2 = = −1 − 𝑖
2(1)
The solution is (-1+i) & (-1-i).

Example (6): If the first root of a quadratic function is -3+2i, what is the second root?
(A) 3+2i
(B) 2+3i
(C) -2-3i
(D) 3-2i
(E) -3-2i

Solution:
From the complex roots, two roots have different sign of complex (i) that means the
answer must be -3-2i
Answer is E.
Example (7): If 𝑓(𝑥) = 𝑥 2 + 𝑘𝑥 + 4, has two real roots ,what is the possible value of k?
(A) 4
(B) 0
(C) -4
(D) -5
(E) 3

Solution:
From the question the function has two real roots means the discriminant is greater
than zero.
𝑎 = 1, 𝑏 = 𝑘, 𝑐 = 4

Then ; √𝑏 2 − 4𝑎𝑐 > 0

√(𝑘)2 − 4(1)(4) > 0

𝑘 2 − 16 > 0
𝑘 2 > 16
𝑘 > 4 𝑜𝑟 𝑘 < −4
The answer is any number greater than 4 or any number less than -4, from the options
(D) is the answer of -5.
Answer is D

Example (8): A function of 𝑓(𝑥) = 𝑥 2 + 6𝑥 + 𝑐 , what is the value of c to make the

function has two equal roots?
Solution:
Has two equal roots means one real root and the discriminant is equal to zero.
𝑎 = 1, 𝑏 = 6, 𝑐 = 𝑐

√𝑏 2 − 4𝑎𝑐 = 0

√62 − 4(1)(𝐶) = 0

36 − 4𝑐 = 0
36 = 4𝑐
Answer is 𝑐 = 9
Perfect & complete Square:
1. (𝑥 − 𝑦)2 = (𝑥 − 𝑦)(𝑥 − 𝑦) = 𝑥 2 − 2𝑥𝑦 + 𝑦 2 “perfect square”

2. (𝑥 + 𝑦)2 = (𝑥 + 𝑦)(𝑥 + 𝑦) = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 “perfect square”

3. 𝑥 2 − 𝑦 2 = (𝑥 − 𝑦)(𝑥 + 𝑦) “Complete square”

Example (9): If 𝑥 2 − 𝑦 2 = 72, and 𝑦 − 𝑥 = 6, what is the value of 𝑥 + 𝑦?

Solution:
𝑥 2 − 𝑦 2 = (𝑥 − 𝑦)(𝑥 + 𝑦)
Given 𝑦 − 𝑥 = 6, 𝑡ℎ𝑒𝑛 𝑥 − 𝑦 = −6:
72 = (−6)(𝑥 + 𝑦)
𝑥 + 𝑦 = −12
Answer is -12

Example (10): If (𝑎 + 𝑏)2 = 60 , and 𝑎2 + 𝑏 2 = 30 , what is the value of 𝑎𝑏?

Solution:
((𝑎 + 𝑏)2 = (𝑎 + 𝑏)(𝑎 + 𝑏) = 𝑎2 + 2𝑎𝑏 + 𝑏 2 = 60
Then;
𝑎2 + 𝑏 2 = 30, 𝑠𝑜
𝑎2 + 2𝑎𝑏 + 𝑏 2 = 30 + 2𝑎𝑏
60 = 30 + 2𝑎𝑏
2𝑎𝑏 = 30
𝑎𝑏 = 15
Answer is 𝑎𝑏 = 15
Sum & Product of roots:
The following rules must be memorized
- Sum of roots:
𝑏
𝑥1 + 𝑥2 = −
𝑎

- Product of roots:
𝑐
𝑥1 . 𝑥2 =
𝑎

Example (11): What is the quadratic equation that has sum of roots equal to -5 and
product of roots equal to 6 ?
Solution:
Quadratic equation is:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Must know a,b and c
Sum of roots:
𝑏
−5 = −
𝑎
Product of roots:
𝑐
6=
𝑎
Assume the leading factor 𝑎 = 1, then:
𝑏
−5 = − , 𝑠𝑜 𝑏 = 5
1
𝑐
6 = , 𝑠𝑜 𝑐 = 6
1

Answer: the quadratic equation is 𝑥 2 + 5𝑥 + 6

Vertex:
The quadratic function has parabolic graph of minimum and
maximum point, this point called vertex.
The vertex point has (x,y) coordinate on xy-plane, to find the
coordinate us the following:
𝑏
𝑋𝑣𝑒𝑟𝑡𝑒𝑥 = −
2𝑎

To find the y- coordinate, substitute the value of 𝑋𝑣𝑒𝑟𝑡𝑒𝑥 in the

equation 𝑓(𝑋𝑣 ).

Axis of symmetry:
The axis of symmetry of a parabola is a vertical line that
divides the parabola into two congruent halves (mirror), it
always passes through the vertex of the parabola.
The equation of the axis of symmetry of the parabola is the x
-coordinate of the vertex (𝑋𝑣 ).

Range of quadratic (𝒚𝒗𝒆𝒓𝒕𝒆𝒙 ):

The range is the possible y-coordinate or the answers of the function.
The quadratic function either it opens upward or downward, if the graph opens
upward that means it has minimum vertex and the possible y-coordinates are equal or
greater than y-vertex.
Minimum vertex:
Range is 𝑦 ≥ 𝑦𝑣𝑒𝑟𝑡𝑡𝑒𝑥
If the graph opens downward and has maximum vertex the range is equal or less
than the y-vertex.
Maximum vertex:
Range is 𝑦 ≤ 𝑦𝑣𝑒𝑟𝑡𝑒𝑥
Example (12): A function 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 5, find the following:
1. X-intercepts (roots).
2. Y-intercept.
3. Vertex.
4. Axis of symmetry.
5. Range.
6. Graph the function (Not SAT requirement).

Solution:
𝑎 = 1, 𝑏 = −6, 𝑐 = 5
1. X-intercept:
Make y=0, 𝑥 2 − 6𝑥 + 5 = 0
Use factoring to find the roots as
𝑥 2 − 6𝑥 + 5 = 0
(𝑥 − 1)(𝑥 − 5) = 0
Then;
𝑥−1=0, 𝑥=1 (1,0)
𝑥 − 5 = 0, 𝑥=5 (5,0)
The x=intercepts are (1,0) 𝑎𝑛𝑑 (5,0).

2. Y-intercept:
Make x=0 , 𝑦 = (0)2 − 6(0) + 5 = 5
Y-intercept is (0,5)

3. Vertex:
𝑏 −6 6
𝑋𝑉𝑒𝑟𝑡𝑒𝑥 = − =− = =3
2𝑎 2(1) 2
𝑌𝑣𝑒𝑟𝑡𝑒𝑥 = 𝑓(3) = (3)2 − 6(3) + 5 = −4
Vertex coordinate is (3, −4)
 4. Axis of symmetry:
Axis of symmetry is equal to x-vertex: x=3

5. Range:
The leading factor (a) is positive means the graph opens upward and the vertex is
minimum so the range is
𝑦 ≥ 𝑦𝑣𝑒𝑟𝑡𝑒𝑥 𝑦 ≥ −4 , [−4, ∞)

6. Graph (Not SAT requirement):