1. • A quadratic function (f) is a function
that has the form as f(x) = ax2 + bx + c
where a, b and c are real numbers and
a not equal to zero (or a ≠ 0).
• The graph of the quadratic function
is called a parabola. It is a "U" or “n”
shaped curve that may open up or
down depending on the sign of
coefficient a. Any equation that has 2
as the largest exponent of x is a
quadratic function.
2. ☺Forms of Quadratic functions:
* Quadratic functions can be expressed in 3
forms:
1. General form: f (x) = ax2 + bx + c
2. Vertex form: f (x)= a(x - h)2 + k (where h and
k are the x and y coordinates of the vertex)
3. Factored form: f(x)= a(x - r1) (x - r2)
3. 1. General form
• Form : f(x) = ax2+ bx+ c
• General form is always written with the x2 term
first, followed by the x term, and the constant term
last. a, b, and c are called the coefficients of the
equation. It is possible for the b and/or c coefficient
to equal zero. Examples of some quadratic functions
in standard form are:
a. f(x) = 2x2 + 3x – 4 (where a = 2, b = 3, c = -4)
b. f(x) =x2 – 4 (where a = 1, b = 0, c = -4)
c. f(x)= x2 ( where a = 1, b and c = 0)
d. f(x)= x2 – 8x (where a = ½, b = -8, c = 0).
4. 2. Vertex Form
• Form: f(x) = a(x - h)2 + k where the point
(h, k) is the vertex of the parabola.
• Vertex form or graphing form of a
parabola.
• Examples:
a. 2(x - 2)2 + 5 (where a = 2, h = 2, and k = 5)
b. (x + 5)2 (where a = 1, h = -5, k = 0)
5. 3.Factored Form
• Form: a(x - r1) (x - r2) where r1 and r2 are
the roots of the equation.
• Examples:
a. (x - 1)(x - 2)
b. 2(x - 3)(x - 4) or (2x - 3)(x - 4)
6. Discriminant
• Quadratic formula: If ax2 + bx + c, a ≠ 0, x =
• The value contained in the square root of the quadratic
formula is called the discriminant, and is often represented
by ∆ = b2 – 4ac.
* b2 – 4ac > 0 → There are 2 roots
x1,2= .
* b2 – 4ac = 0 → There is 1 real root,
x = -b/2a.
* b2 – 4ac < 0 → There are no real
roots.
7. Using Quadratic Formula
• A general formula for solving quadratic equations, known as
the quadratic formula, is written as:
• To solve quadratic equations of the form ax2+ bx+ c,
substitute the coefficients a, b, and c in to the quadratic
formula.
1. Exercise: Solve 4x2 – 5x + 1 = 0 using the quadratic
formula
∆ = b2 – 4ac= 52 – 4(4)(1) = 25 – 16 = 9 => = = ± 3
x
x1 = 1 or x2 = 1/4
8. Using Factoring
• Convert from general form, f(x) = ax2 + bx + c to factored form,
a(x - x1) (x - x2) :
+ Example 1: Solve x2+ 2x = 15 by factoring
=> x2 + 2x – 15 = 0 (General form)
(x + 5)(x - 3) = 0 (Factoring form)
x + 5 = 0 or x – 3 = 0
x = -5 or x = 3
→ Thus, the solution to the quadratic equation is x = -5 or x = 3.
+ Example 2: Solve x2 + 5x – 9 = -3 by factoring
=> x2 + 5x – 6 = 0 (General form)
(x – 1)(x + 6) = 0 (Factoring form)
x – 1 = 0 or x + 6 = 0
x = 1 or x = -6
→ Thus, the solution to the quadratic equation is x = 1 or x = -6.
9. Using Completing the Square
• Converting from the general form f(x) = ax2+ bx+ c to a
statement of the vertex form f(x)= a(x - h)2 + k.
• When quadratic equations cannot be solved by factoring,
they can be solved by the method of completing the squares.
• Example: Solve x2 + 4x – 26 = 0 by completing the square
• (x2 + 4x + 4 – 4) – 26 = 0 (General
form)
• (x + 2)2 – 4 – 26 = 0
• (x + 2)2 – 30 = 0 (Vertex form)
• (x + 2)2 = 30
• x + 2 =
• x = - 2
• x = - 2 + or x = - 2 -
10. Application to higher-degree
equations
• Example: x4 + 4x2 - 5 = 0
– The equation above can be written as:
– (x2)2 + 4(x2) - 5 = 0
– (Quadratic function with exponent x = 2)
– Solve: ./ Substitute x2 = P
P2 + 4P - 5 =0
(P + 5)(P - 1) = 0
P = -5 or P = 1
./ Re-substitute P = x2
=> P = -5 = x2 = no roots
=> P = 1 = x2 => x = ± 1
11. The Graph of Quadratic Equation
• The graph of quadratic equation in the form f(x) = ax2+ bx+ c is a
parabola. The parts of the graph of the parabola are determined by the
values of a, b, and c.
• The most meaningful points of the graph of a parabola are:
1. x-intercepts: The x-intercepts, if any, are also called the roots of the
function. They are meaningful specifically as the zeroes of the
function, but also represent the two roots for any value of .
2. y-intercept: The importance of the y-intercept is usually as an
initial value or initial condition for some state of an experiment,
especially one where the independent variable represents time.
3. Vertex: The vertex represents the maximum (or minimum) value
of the function, and is very important in calculus and many natural
phenomena.
12. X-intercepts
• The x intercepts of the graph of a quadratic function f given by y = ax2 + bx + c
• The x-intercepts are the solutions to the equation ax2 + bx + c = 0
• The x-intercept in the equation f(x) = ax2 + bx + c, can be found in basically
two ways, factoring or the quadratic formula.
•
* Factoring: If f(x) = y = ax2 + bx + c can be factored into the form y = a(x –
r1)(x – r2) , then the x-intercepts are r1 and r2 .
• Example: y = x2 - 3x - 18
=> 0 = x2 – 3x – 18 (Set y = 0 to find the x – intercept)
0 = (x – 6)(x + 3) (Factor)
x – 6 = 0 or x + 3 = 0
x = 6 or x = -3 ( Solved the equation => x – int. = (6, 0) ; (-3,
0) )
• * Quadratic formula: For any function in the form y = ax2 + bx + c, x-
intercepts are given by:
( , 0)
• Example: y = 2x2 + 5x + 3 (a = 2; b = 5; c = 3 )
=> 0 = 2x2 + 5x + 3 (Set y = 0 to find x – int. )
x = ( Using Quadratic formula to solve the equation)
x = -5 ±1/4 => x = -19/4 or x = -21/4
(Simplified the equation => x – int. = (-21/4, 0) or (-19/4, 0)).
13. Y-intercept
• The y intercept of the graph of a quadratic function is
given by f(0) = c or y-intercept = (0,c).
• Example:
Find the y intercept of the following equations:
– A) x2 + 2x + 26
– B) 16x2 – 3x – 59
– C) 10x2 + 5x – 10/7
* Solution:
A) Substitute 0 for x as f(x) = f(0) = x2 + 2x + 26
=> f(0) = c = 26.
The y intercept is at (0, 26).
B) Same method as A). f(0) = -59.
The y-int. is (0, -59)
C) f(0) = -10/7.
The y-int. is (0, -10/7)
14. The Vertex
• The vertex can be found by completing the
square, or by using the expression derived from
completing the square on the general form.
• Formula given by: y = ax2 + bx + c So the
vertex is
• Example: h = -12x2 + 168x – 38
• => -b/2a = -168/2(-12) = 7 (applied the formula
to find x-coordinate of vertex of the quadratic
equation).
• Substitute for x (x = 7) into the original
equation. We have that: h = -12(7)2 + 168(7) –
38 = 550
• Thus, the vertex of the equation is (7, 550).
15. Graphing Parabola
• There are several steps to do before we sketch a graph of
parabola of the quadratic equation.
• The most easiest way to sketch the graph is from vertex
form.
• Steps:
1. We need to check whether the graph is concave up if a > 0
(U- shaped) or concave down if a < 0(n-shaped).
2. We need to find the vertex of the equation by the formula
x =-b/2a or by the vertex form.
3. We need to find the values of x and y intercepts.
+ Substitute x = 0 into the equation then find the
coordinate of y value.
+ Let y = 0 then find the values of x.
16. Graphing Parabola
• Graph the function f(x) = 2x2 + 8x +7
Solution:
1. Write the function f(x) = 2x2 + 8x + 7 in the form f(x) = a (x - h)
2 + k by completing the square. (h, k)
f (x) = 2x2 + 8x + 7 => f (x) = 2 (x2 + 4x) + 7
=> f (x) = 2 (x2 + 4x + 4) + 7 – 8
=> f (x) = 2 (x + 2)2 – 1
3. The function f (x) = 2x2 + 8x + 7 has vertex at (-2, -1) with a
horizontal shift of 2 units to the left and a vertical shift of one
unit downward. Also since a = 2 > 0 then the graph is concave
up.
4. The x-intercept of the function are determined by letting
f (x) = 2 (x + 2)2 – 1 = 0 and solving for x as illustrated below:
17. Continued
f (x) = 2 (x + 2)2 – 1 = 0
=> (x + 2)2 = ½
=> x + 2 = ± √½ = ± 1/√2
=> x = -2 ± (1/√2)
4. The y-intercept is f (0) = 2 (0 + 2)2 – 1 = 7
* The graph of f (x) = 2x2 + 8x +7 is
illustrated below:
