Unit 6B Review

• The graph has a y-intercept at (0, 0)

• The coordinates of the vertex is (1, -1)

• The graph of the function opens downward

• The x-interepts (zeros/roots/solutions) are at (0,0) and (2,0)

The graph for function h is pictured, SELECT ALL that could define function h (***USE DESMOS***):

• g(x) = (x-2)(x-4)

• g(x) = (x+2)(x+4)

What are the x- intercepts?

• x= 0 and x= -4
• x= 0 and x= 4
• y= 0
• x= 2

What is the y-intercept?

• (2, -2)

• (4,0)

• (0,0)

• (-1,2)

What is the vertex?

• (2, 4)
• (3, -1)
• (0, 8)
• (4, 2)

What is the y-intercept?

• (2, 0)

• (3, -1)
• (0, 8)
• (4, 0)

• The graph has an x-intercept at (c,0)

• When a < 0 the graph opens downward

• If b=0, then the vertex is on the y-axis

• The y-intercept of the graph is at (0, c)

• The graph has two x-intercepts.

A quadratic function f is defined by f(x)= (x+4)²-5

What is the vertex of the function? (***CHECK WITH DESMOS***)

• (-3, -5)
• (-4, -6)
• (-5, -4)
• (-4, -5)

A quadratic function f is defined by f(x)= (x+4)²-5

Does the vertex of the function represent the minimum value or maximum value of the function? (***CHECK WITH DESMOS***)

• minimum value

• maximum value

A quadratic function f is defined by f(x)= (x+4)²-5

If you were to shift this graph 6 units up from where it is now, what would be the equation

represented by the new graph? (***CHECK WITH DESMOS***)

• f(x) = (x+4)² + 6

• f(x) = (x+4)²- 11

• f(x) = (x+4)² + 3

• f(x) = (x+4)² + 1