Quadratic functions are defined by: y = f(x) = ax2 +bx + c = 0

The graph of a quadratic function is a parabola.

The most basic quadratic function is: y = x2, where a = 1, b = 0 and c = 0.

approachesxfapproachesxas )(,

approachesxfapproachesxas )(,

xasxwriteWe 2:

00min2 xatofvalueimumahasxy

Note: We use the graph of f(x) = x2 to determine the graphs of other quadratic functions using a shifting technique.

2)1()(:1 xxgofgraphtheSketchEx

1st: identify the function. Quadratic. We now know that it will be similar to y = x2

2nd: Note the differences. There is a 1 subtracted from the x before the squaring of x. This will cause a horizontal shift.

When the addition/subtraction occurs before the squaring of x, we do its opposite: subtraction to the right, addition to the left.

The graph of y = x2 will get shifted one unit to the right.

Horizontal shifts of graphs:

1. The graph of y = f(x-c) is the graph of y = f(x) shifted c units to the right.

2. The graph of y = f(x + c) is the graph of y = f(x) shifted c units to the left.

These are called horizontal translations.

Standard form of a quadratic equation: f(x) = ax2

Recall: We use the technique of completing the square on circles to find: - center - radius

We use this same technique to find the horizontal/vertical translations and vertexof quadratic equations.

Ex 2: Sketch the graph of 32)( 2 xxxf

To see the shifts, we must complete the square on the x term.

3))(1()2()( 2 xxxf

Note: Before you can complete the square, the value of “a” must be 1.

2)1()(

3)1)(1()12()(2

2

xxf

xxxf

We now see that the graph of the parent function is shifted 1 unit to the rightand 2 units up.

The vertex of our parabola is: (1, 2)

Vertical Shifts of Graphs

Let c > 0.

1. The graph of y = f(x) + c is the graph of y = f(x) shifted upward c units.2. The graph of y = f(x) – c is the graph of y = f(x) shifted downward c units.

The graph of a quadratic equation

1. The graph opens upward when a > 0, and opens downward when a < 0.2. The greater the magnitude of a, the steeper the graph and the narrower the opening.

xasxfthenaif )(,0.)3

xasxfthenaif )(,0

Ex 3: Sketch the graph of 17122)( 2 xxxf

Complete the square on the quadratic

17)122()( 2 xxxf

17))(2()6(2)( 2 xxxf

17)9)(2()96(2)( 2 xxxf

1)3(2)( 2 xxf

The vertex is: (-3, -1)This graph will be the graph of the parent function vertically elongated by a factor of 2, shifted three units to the left and 1 unit down.

Ex: 4 Sketch the graph of 3231

)( 2 xxxf

Solution: Complete the square, but first factor out (-1/3).

3)(31

)6(31

)( 2

xxxf

3)9(31

)96(31

)( 2

xxxf

33)3(31

)( 2 xxf

6)3(31

)( 2 xxf

The vertex is: (3, 6)

The graph of the parent function ismultiplied by a factor of (-1/3), shifted to the right 3 units and shifted up 6 units.

Ex 5: A company finds that it can sell x units per day when the price is P(x)=100 – 0.05x, for x between 250 and 800. The cost of operating the company, regardless of the number of items produced, is $4000 per day, and the production cost per unit when x items are produced is 60 – 0.01x. Determine the number of units that should be produced each day to maximize profit, and find its maximum profit.

Solution: Profit is calculated by taking the revenue minus the cost.P(x) = R(x) – C(x).

C(x) = 4000 + x(60 – 0.01x)

201.0604000)( xxxC

)05.0100()( xxxR 205.0100)( xxxR

)01.0604000(05.0100)( 22 xxxxxP

Cost

Revenue

40004004.0)( 2 xxxP

40004004.0)( 2 xxxP

Now, complete the square to find the vertex. The vertex will tell the maximumprofit and at what number of items that need to be produced to achieve thatmaximum profit.

6000)500(04.0)( 2 xxP

The vertex is: (500, 6000). This means that the maximum profit that can be made in a day is $6000 if 500 items are produced.