Download Section 2.1

Quadratic Functions
A regular parabola comes in two forms (we’ll take another look at parabola as a conic section later in the course):
1) y = ax2 + bx + c
general form
where a, b, c are some constants
2) y = a(x – h)2 + k
vertex (standard) form
where (h, k) is a vertex
Quick review on transformations (College Algebra):
To graph y = a(x – h)2 + k
The graph of y = x2 (parabola with vertex at the origin) is
1) reflected about x-axis if a is negative
2) |a| > 1, stretched (the graph appears narrower)
3) |a| < 1, compressed (the graph appears wider)
4) shift h units
a) right if h is positive (‘ – ‘ in the original equation)
b) left if h is negative (‘+ ‘ in the original equation)
5) shift k units
a) up if k is positive (‘ + ‘ in the original equation)
b) down if k is negative (‘ – ‘ in the original equation)
1) Graph the parabola.
y = -2(x – 1)2 + 3
a = -2, h = 1, k = 3
The graph is reflected about x-axis (a is negative), stretched (|a| >1), shifted 1 unit to the right
and 3 up.
Graph:
2) Graph the parabola. Describe all transformations.
y=
1 2
x 1
3
a=
1
, h = 0, k = 1
3
The graph is compressed (|a| < 1), shifted 1 unit up.
Graph:
Another method of graphing y = ax2 + bx + c
Find a, b, c.
1) If a > 0, the parabola opens up
If a < 0, the parabola opens down
2) Find vertex:
x = -b/2a
y = f(-b/2a)
vertex: (
b
b
, f(
))
2a
2a
3) Find y – intercept: (0, c)
4) Find x – intercepts:
2
ax + bx + c = 0
Factor or use Quadratic Formula: x
b
b2
2a
1) 2 x – intercepts if b2 – 4ac > 0
2) 1 x – intercept if b2 – 4ac = 0, and it’s the vertex
3) no x – intercepts if b2 – 4ac < 0
5) Graph the parabola. Find the point symmetric to y-intercept if necessary.
1) Graph the parabola by following the method above.
y = x2 – 2x + 4
4ac
a = 1, b = -2, c = 4
1) the parabola opens up
2) Vertex: x = −
( )
=1
y = f(1) = 12 – 2(1) + 4 = 3
Vertex: (1, 3)
3) y – int: (0, 4)
4) x2 – 2x + 4 = 0 Let’s check the discriminant before using the Quadratic Formula.
b2 – 4ac = (-2)2 – 4(1) (4) = 4 – 16 = -12
There are no x-intercepts.
5) Graph:
There are some nice applications of parabolas, for example, maximizing revenue in the business
situation, and finding optimal price (the price that gives maximum revenue).
1) A demand equation p = 200 – 4x, where p is the price per unit and x is the number of units
produced and sold (the demand). Find x for which the revenue is maximum, maximum revenue
and the optimal price. Recall that R(x) = x *p.
Solution: First, let’s find R(x).
R(x) = x (200 – 4x) = 200x – 4x2 = -4x2 + 200x
To find x that maximizes revenue and maximum revenue, simply find the vertex for R(x).
a = -4, b = 200
xmax = -200/(2(-4)) = 25 units
Rmax = -4 (25)2 + 200 (25) = $2500
This is maximum revenue.
To find optimal price, we’ll plug the found values into R(x) = x*p.
2500 = 25p
p = $100
The optimal price to charge for this unit is $100.
2) Minimizing cost. The total cost for a product is C = 200 – 50x + x2, where x is the number of
units produced. Find the value of x for which the total cost is minimum.
Solution: Since Cost is a quadratic function, then the minimum of a function is a vertex of the
parabola.
To find vertex: a = 1, b = -50
x = b/2a = - (-50)/2(1) = 25 units
If 25 units are produced, then the cost is at a minimum.