Download 4.3 Quadratic functions and their properties A quadratic function is a

4.3 Quadratic functions and their properties
A quadratic function is a function defined as f(x) = ax2 + bx + c, a 0
Domain: the set of all real numbers
x-intercepts: Solutions of ax2 + bx + c = 0
y-intercept: y = f(0) = c
Graphs of f(x) = ax2
The graph of y = ax2 is a parabola with vertex at (0,0).
a>0
a<0
Remarks: - Note that the larger |a|, the narrower the parabola
- The vertical line through the vertex is the axis of symmetry for the parabola
We can graph a quadratic function using transformations.
Example: Graph, using transformations, f(x) = 2(x-1)2 – 3
The order of transformations is following
1) y = x2
basic function
2) y = 2x2 stretch by a factor of 2
3) y = 2(x-1)2 shift right by 1
4) y = 2(x-1)2 – 3
shift down by 3
Note that the vertex is now at (1,-3)
Any quadratic function f(x) = ax2 + bx + c can be written in the form f(x) = a(x-h)2 + k and then graphed using
transformation
Example: Graph, using transformations, f(x) = 2x2 - 4x - 1
(a) We must first write this function in the form f(x) = a(x-h)2 + k
(i)
(ii)
(iii)
(iv)
Enclose the terms that contain x in the parenthesis
f(x) = (2x2 – 4x) - 1
Factor out coefficient of x2 from the group in the parentheses
f(x) = 2(x2 -2x) – 1
Complete the expression in parentheses to a perfect square and add the opposite of the added
number to the whole expression
f(x) = 2(x2 – 2x +1) - 1 + 2(-1)
Write the expression in parentheses as a square, add the numbers
f(x) = 2(x-1)2 – 3
(b) graph f(x) = 2(x-1)2 – 3 ( see the previous example)
We can perform the same operations on the general formula f(x) = ax2 + bx + c to obtain
b 
4ac  b 2

f ( x )  a x 
 
2a 
4a

2
Therefore, we can conclude the following
_ the graph of a quadratic function f(x) = ax2 + bx + c is a parabola that opens up when a > 0 and opens down
when a < 0
 b 4ac  b 2   b
 b 
    , f    
_ the vertex of the parabola is at   ,
4a   2a  2a  
 2a
_ the axis of symmetry is the vertical line x = 
b
2a
Using this information and the fact that
_ the y- intercept is (0,c)
_ the x-intercepts , if any, are the solutions of ax2 + bx + c = 0,
we can graph the parabola without completing the square and performing transformations
Example: Graph f(x) = 2x2 - x – 1
(i)
Determine whether the parabola opens up or down:
Since a = 2 > 0, the parabola opens up
(ii)
Find the vertex
b
1 1
x-coordinate of the vertex = 


2a
22 4
2
1 1
9
 b 
1
1 1
y-coordinate of vertex = f     f    2      1    1  
8 4
8
 2a 
4
4 4
1 9
vertex =  , 
4 8
(iii) Find the equation of the line of symmetry:
It is a vertical line passing through the vertex, so the equation is x = ¼
(iv)
Find the y-intercept:
y = f(0) = c = -1
y-intercept is (0, -1)
(v)
Find the x-intercepts, if any, by solving the equation f(x ) = 0:
2x2 – x -1 = 0
(2x+1)(x-1) = 0
x = -1/2 or x = 1
x-intercepts are (-1/2, 0), (1,0)
(vi)
Use (i) – (v) to graph the parabola
Since the graph of any quadratic function is a parabola, the vertex is either the lowest (when a >0)
or the highest (a < 0) point on the graph. The highest point on the graph of any function is called
the maximum and the lowest point is called the minimum.
If a point (a,b) is the maximum for a function f, then f(x) < b for all values x (in the domain of f)
and b is called the maximum value of f
If a point (a, b) is the minimum for a function f, then f(x) > b for all values x (in the domain of f)
and b is called the minimum value of f
Therefore, a quadratic function has the maximum/ minimum at x = 
b
and the maximum/
2a
 b 
minimum value is f   
 2a 
Example: Determine whether f(x) = -2x2 -3x + 5 has the maximum or minimum and find this
value.
Since a = -2, the graph of y = f(x) is a parabola that opens down, and therefore the vertex is the
highest point on the graph. f(x) has the maximum at the vertex, that is when
b
3
3

  . The maximum value is
x= 
2a
2  (2)
4
 b 
f   
 2a 
9 9
49
 3
 3
  3
f     2    3
5    5
8 4
8
 4
 4
 4 
2
4.4 Quadratic models
In this section you will be asked to find the smallest/largest value of some quantity Q. The steps
that must be taken are
(i)
Find the formula for the quantity Q. Base your answer on known formulas and the
information given in the problem
(ii)
Identify the formula for Q as quadratic; based on the value of a determine whether Q has
the minimum or maximum and find its value
4.5 Inequalities involving quadratic functions
A quadratic inequality is an inequality ax2 + bx + c > 0 (or > 0 , < 0 , < 0).
Since ax2 + bx + c is a polynomial, we can solve this inequality as any other polynomial inequality.
Example: Solve x2 + x – 1 > 0
(i)
solve the equation: x2 + x – 1 = 0
We must use the quadratic formula: x 
 b  b 2  4ac  1  1  4(1)(1)  1  5


2a
2
2
(ii) plot the solutions on the number line and use test points to determine the sign of x2 + x -1 in each interval
Positive
negative
positive
1 5
2
1 5
2


1 5   1 5


Therefore, x2 + x -1 > 0 when x is in   ,
,


 

2
2

 

Another way to solve a quadratic inequality is to use the properties of the quadratic function and functions in general. If
the graph of a function f is above x-axis, then f(x) > 0 (above x-axis are points with y-coordinate greater than 0, and the
graph is the set of points (x, f(x))); when the graph is below x-axis, f(x) < 0; at points where the graph crosses or
touches x-axis, f(x) = 0.
Example: The graph of a function f is given below. What is the solution of the inequality f(x) > 0?
We are looking for x for which f(x) > 0, which means we are looking for x for which the graph is above x-axis.
1) Identify parts of the graph that are above x-axis
2) Determine values of x that produce these parts and write them in the interval notation.
The graph is above x- axis for x in (-3, 1) and (4, +). Therefore, the solution of f(x) > 0 is
(-3, 1)  (4, +).
Remark: Include the x-intercepts only when the inequality is < or > .
We can apply this technique for quadratic functions, since we can quickly sketch its graph.
Example: Solve 6x2 < 6 + 5x
(i)
Write the inequality in the standard form (0 on the right)
6x2 – 5x – 6 < 0
(ii)
Sketch the graph of f(x) = 6x2 – 5x -6 using only the x-intercepts and the fact that the graph is a parabola
that opens up (a > 0)
x-intercepts: 6x2 – 5x -6 = 0
(2x-3)(3x+ 2) = 0
x = 3/2 or x = - 2/3
Since we want 6x2 -5x- 6 < 0 (or f(x) < 0), we have to identify the part(s) that is below x-axis. This part is obtained
when x is in (-2/3, 3/2). Since the inequality is <, we will include the x-intercepts. Therefore, the solution is [-2/3, 3/2]
Remark: If there are no x-intercepts, then the graph will be entirely above or below x-axis. You will need vertex to
sketch the graph. If the graph is entirely above the x-axis, then f(x) > 0 for all real numbers, and f(x) < 0 has no
solutions. If the graph is entirely below the x-axis, then f(x) < 0 for all real numbers, and
f(x) > 0 has no solutions.