1. Quadratic Functions and their graphs Lesson 1.7
A quadratic function- f(x) = ax2 + bx + c where a=/ 0
The graph of a quadratic function is called a parabola
Two very special parts
of a parabola:
Vertex: The ‘turning’ point
It is either a maximum or
minimum.
Axis of symmetry: A vertical
Line that passes through the
Vertex.

2. The axis of symmetry can always be found by
calculating: x = - (b) 2a
Vertex: ( - b , f( - b) )
2a a
Discriminant: If b2 – 4ac > 0 Parabola crosses x-axis
twice. There will be two
x-intercepts.
b2 – 4ac = 0 Parabola is ‘tangent’ to
x-axis. There is only
one x-intercept.
b2 – 4ac < 0 Parabola never crosses
the x-axis.
No x-intercepts.

3. The axis of symmetry is a vertical line midway between
the x-intercepts. Therefore it is
the ‘average’ of the x-intercepts.
Example: Find the intercepts , the axis of symmetry ,
and the vertex , of this parabola.
y = (x+4)(2x – 3)
To find x-intercepts: Replace y with 0.
0 = (x+4)(2x – 3)
Set each factor = 0
x + 4 = & 2x – 3 = 0
x = x = 3/2
Since the x-intercepts have already been found
Find the average of these to find the axis.
x = /2 = = -1.25
Now  Vertex = (-1.25, f(-1.25))
(-1.25, )
Sketch the graph

4. Example: Sketch the graph of the parabola. Label the
intercepts, the axis of symmetry, and the
vertex.
y = 2x2 – 8x + 5
1st find the Axis of Symmetry
X = -(-8) = 8 = 2
2(2)
Now find the vertex:
V = (2, f(2)) = (2,- 3)
On this one, since b2 – 4ac = 44 ,
Which is not a perfect square, this can-
Not be factored  use the quadratic
Formula to find x intercepts.
X = 0.78 & 3.22
If you remember ‘c’ is always the y-intercept
Sooo  y-int = 5.
Draw the ‘vertical axis’ at x = 2, Plot the vertex at (2,-3)
Estimate the x-intercepts at 0.78 and 3.33,
Plot the y-intercept at y = 5 and plot its symmetry
point at (4,5) and sketch the parabola!
If the equation can be written in the form of :
y = a(x – h)2 + k vertex -- (h,k)
axis of sym x = ‘h’

5. If the equation can be written in the form of :
y = a(x – h)2 + k vertex -- (h,k)
axis of sym x = ‘h’
Example: a) Find the vertex of
the parabola y = - 2x2 + 12x + 4
by completing the square.
y = -2x2 + 12x + 4 = st subtract 4 from both sides
y – 4 = -2x2 + 12x factor out a – 2 from both ‘x’terms
y – 4 = -2(x2 – 6x ) complete the square inside the parentheses
b = -6/2 = (-3)2 = 9  now add 9 inside the ( ) but add -2(9) or -18 to other side
y – 4 – 18 = -2(x2 -6x +9)  change to this look
y – 22 = -2(x – 3) Now add 22 back to the right side
y = -2(x – 3) line up y = a(x – h)2 + k
y = a(x – h)2 + k Identify h = 3, k = 22 so
vertex = (h,k) = (3,22)

6. Example: b) Find the ‘x’ and ‘y’ – intercepts.
y – intercept can be found from the given equation
‘c’ = y –intercept so y-intercept = 4
to find x-intercepts: let y = 0 and get:
0 = -2(x – 3)2 + 22
- 22 = -2(x – 3)2
11 = (x – 3)2
+ √11 = x – 3
3 + √11 = x
Example: Find the equation of the quadratic function
‘’f’ with f(-1) = - 7 and a maximum value of
f(2) = -1
f(-1) =  (-1,-7)
A Maximum value at f(2) = -1 means (2,-1) is the vertex
(h,k)

7. so using h = 2 and k = - 1 gives us this working format:
y = a(x – 2)2 + (-1)
 using the other point given (-1,-7) for x and y gives us:
- 7 = a(-1 – 2)  solve for ‘a’
- 6 = a(-3)2
- 6 = 9a  divide by 9
- ⅔ = a
So putting it all together  y = -⅔(x – 2)2 - 1
b) Show that the function ‘f’ has
no x-intercepts.
The parabola must open downward since a = - ⅔ and since
the vertex is located below the x-axis at (2,-1) it cannot
cross the x-axis!

8. Example: Where does the line y = 2x + 5
intersect the parabola y = 8 – x2
Check out the solution for example 3
found on page 40!
(Show both the algebraic and graphical approach)

9. Example: Find an equation of the function whose
graph is a parabola with x-intercepts
‘1’ and ‘4’ and a y-intercept of ‘- 8’
Look over example 4 found on page 40.
Homework: page 40 CE #1-6 all;
page 41 WE #1-25 left column