4.1 – 4.3 Review

Sketch a graph of the quadratic.y = -(x + 3)2 + 5Find: Vertex(-3, 5)Axis of symmetryx = -3y - intercept (0, -4)x - intercepts

Describe how this quadratic has been transformed in comparison to the “classic” parabola.

The parabola is the same shape as the classic but opens down. The parabola is shifted left 3 units and up 5 units.

( 5 3,0) and ( 5 3,0)

Sketch a graph of the quadratic.y = 2x2 -20x + 50 y = 2(x – 5)2+0

Find: Vertex(5,0)Axis of symmetryx=5y - intercept (0,50)x - intercepts(5, 0)Describe how this quadratic has been transformed in

comparison to the “classic” parabola. The parabola is shifted right 5 units and stretched vertically by a factor of 2 (is narrower than the classic).

Rewrite in vertex form. Then, identify the vertex for each. (Complete the Square!!!)

1. f(x) = x2 -20x + 4 3. f(x) = 2x2 -2x + 5

2. f(x) = 5x2 +20x

2

2

( ) ( 20 100) 4 100

( ) ( 10) 96

f x x x

f x x

2

2

2

( ) 2( ___) 5 ____

1 1( ) 2( 1 ) 5 2

4 4

1 9( ) 2( )

2 2

f x x x

f x x x

f x x

2

2

2

( ) 5( 4 ___) ____

( ) 5( 4 4) 5(4)

( ) 2( 2) 20

f x x x

f x x x

f x x

Find the real zeros. 1. y = x2 +7x – 8 3. y = -4(x-9)2 + 4 0 = (x+8)(x-1)

x = -8, 1

2. y = 2x2 - 28x + 980 = 2(x2-14x +49)0 = 2(x-7)2

Do you always factor to find the zeros? No see #3

The real zeros of the function are also called the solutions of the equation and the

x – intercepts of the graph.

0 =-4(x-9)2 + 4-4 = -4(x-9)2

1 = (x-9)2

±1 = (x – 9)x = 1+9 , -1 +9x= 10, 8

Use the calculator to find the zeros and the vertex of each. *** adjust the WINDOW as needed to see the graphs and indicate below!!

1. f(x) = -x2 -10x + 4 2. f(x) = 2x2+8x-4

Use calculator graphing features.

Window Setting: #1 x Min____ x Max___ #2 x Min____ x Max___

y Min____ y Max ____ y Min____ y Max ____

In the “calculate” menu (2nd Trace), the vertex can be found by calculating the maximum or minimum.