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• 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.
☺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)
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).
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)
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)
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.
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
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.
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 -
	
   	
   	
  	
  
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
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.
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)).
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)
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).
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.
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:
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:
Quadraticfunctionpresentation 100127142417-phpapp02
Helpful Links
•  h#p://www.douglas.bc.ca/services/learning-­‐
centre/pdf/math/
MA7_30_QuadraBc_EquaBons_and_FuncBons
.pdf	
  
•  h#p://www.cimt.plymouth.ac.uk/projects/
mepres/book9/y9s17ex.pdf	
  
•  h#p://www.tech.plym.ac.uk/maths/
resources/PDFLaTeX/quad_graphs.pdf	
  
•  h#p://www.neufeldmath.com/ss/Plus/
Graphing%209.pdf	
  

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Quadraticfunctionpresentation 100127142417-phpapp02

  • 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:
  • 19. Helpful Links •  h#p://www.douglas.bc.ca/services/learning-­‐ centre/pdf/math/ MA7_30_QuadraBc_EquaBons_and_FuncBons .pdf   •  h#p://www.cimt.plymouth.ac.uk/projects/ mepres/book9/y9s17ex.pdf   •  h#p://www.tech.plym.ac.uk/maths/ resources/PDFLaTeX/quad_graphs.pdf   •  h#p://www.neufeldmath.com/ss/Plus/ Graphing%209.pdf