Pc Functions Quads

f ( x) = x2

f ( x) =

f ( x ) = 3x 2

f ( x ) = ax

1 2 x 3

If a < 0 the parabola opens down and the larger the a the “narrower” the graph and the smaller the a the “wider” the graph.

2

If a > 0 the parabola opens up and the larger the a value the “narrower” the graph and the smaller the a value the “wider” the graph.

1 f ( x) = − x2 3

f ( x ) = − x 2 f ( x ) = −3x 2

The graph of this function is a parabola

f ( x ) = a( x − h ) + k 2

Determines whether the parabola opens up or down and how “wide” it is

horizontal shift, moves graph horizontally by h

vertical shift, moves graph vertically by k

Thisneed We will factor to algebraically into (x+3)(x+3) so manipulate we can express this to it as look something like the Let’s look squared at a above. form and combine We’ll do this the by -1 and quadratic -9 function completing on the end. the square. and see if we can 2 graph it. 9 9

f ( x) = x − 6x −1 2

f ( x ) =f ( xx ) − ) −−101 − ____ = 6( xx −+3___ 2

Subtract it here to keep things equal Add a number here to make a perfect square

(can’t add a number without compensating for it and we don’t want to add it to the other side because of function notation)

down 10

f ( x ) = ( x − 3) − 10 2

right 3

( )

We started with f x = x 2 − 6 x − 1 and completed the square to get it in the format to be able to graph using transformations. We can take a general quadratic equation and do this to find a formula for the vertex. This is done in your book at the bottom of page 295 by the number 2. What we find from doing this is on the next slide.

f ( x ) = ax + bx + c 2

Let’s try this on the one we did before: 2

f ( x ) =1x − 6 x − 1

The x value of the vertex of the parabola can be found by computing − b

2a

− (-6) b x value of vertex = 2a(1)

=3

The vertex is then at (3, -10)

The y value of the vertex of the parabola can be found by substituting the x value of the vertex in the function and finding the function value.

y value of vertex = ( 3) − 6( 3) − 1 = −10 2

Let’s plot the vertex:

(0, -1)

(6, -1)

Since the a value is positive, we know the parabola opens up. The parabola will be symmetric about a vertical line through the vertex called the axis of symmetry.

ff( (xx) ) ==0x −−66( 0x )−−11 22

(3, -10) Let’s find the y intercept by plugging 0 in for x. So y intercept is (0, -1) We can now see enough to graph the parabola

The graph is symmetric with respect to the line x = 3 so we can find a reflective point on the other side of the axis of symmetry.

Let’s look at another way to graph the parabola starting with the vertex: We could find the x intercepts of the graph by putting f(x) (which is the y value) = 0

f0( x=) x= x− 6−x6−x1− 1 2 2

(3, -10)

This won’t factor so we’ll have to use the quadratic formula.

− ( − 6 ) ± ( − 6 ) − 4(1)( − 1) − b ± b − 4ac = x= 2(1) 2a 6 ± 40 = ≈ 6.2 and − .2 So x intercepts are (6.2, 0) and (- 0.2, 0) 2 2

2

A mathematical model may lead to a quadratic function. Often, we are interested in where the function is at its minimum or its maximum. If the function is quadratic the graph will be a parabola so the minimum (if it opens up) will be at the vertex or the maximum (if it opens down) will be at the vertex.

We can find the x value of the vertex by computing

b − 2a

We could then sub this value into the function to find its minimum or maximum value.

DEMAND EQUATION The price p and the quantity x sold of a certain product obey the demand equation:

1 p = − x + 200, 0 ≤ x ≤ 400 2 Express the revenue R as a function of x. Revenue is the amount you bring in, so it would be how much you charge (the price p) times how many you sold (the quantity x)

This is the real world domain. The equation doesn’t make sense if the quantity sold is negative (x < 0) and it doesn't make sense if the price is negative (if x > 400)

R = xp

1 2  1  R = x − x + 200  = − x + 200 x 2  2 

1 2 R = − x + 200 x 2

This is a quadratic equation and since a is negative, its graph is a parabola that opens down. It will have a maximum value then at the y value of the vertex.

What is the revenue if 100 units are sold?

1 2 R = − (100 ) + 200(100 ) = $15,000 2 b

What quantity x maximizes revenue? x = −

2a

=−

200 = 200  1 2 −   2

Since the revenue function is maximum at the vertex, we'll want to find the x value of the vertex to answer this.

1 2 ( ) ( ) f 200 = − 200 + 200( 200 ) What is the maximum revenue? 2 This would be the y value of the vertex

= $20,000

DEMAND EQUATION The price p and the quantity x sold of a certain product obey the demand equation:

1 p = − x + 200, 0 ≤ x ≤ 400 2

1 2 R = − x + 200 x 2

What price should the company charge to receive maximum revenue? Since we just found that the quantity to achieve maximum revenue was 200, we can substitute this in the price equation to answer this question.

1 p = − ( 200 ) + 200 = $100 2