Zerossss
When finding a zero of the function, what is it that you are looking for?
x-intercept
Rewrite (x-3)^2+8 into the standard form
x^2-6x+17
Name the x-intercepts. y=2(x-2)(x+6)
(2,0) and (-6,0)?
Name the vertex y= (x-3)^2 +4
(3,4)
Find the y-intercept for 4x^2+3x-7
(0,-7)
Its' x-coordinate is ALWAYS a zero.
y-intercept
Given the standard form: ax^2+bx+c, what happens when variable a<0?
Parabola opens downward.
Name the x-intercepts Y=(x)(x-2)
(0,0) and (2,0)?
y= (x+1)^2 - 8
(-1, -8)
Find the y-intercept for y = (x+2)(x-2)
(0,-4)
Its' y-coordinate is always a zero.
x-intercept
A ball is being thrown up in the air by a child. Given what you know about quadratic functions, when fining time when the object hits the ground, what is it that you are looking for on the graph?
the x-intercept.
Name the x-intercepts. y=x^2 - 4
(-2,0) and (2,0)?
y=(x+4)(x-4)
(0,-16)
Is f(x)=5x-5 a quadratic function? Why yes/no?
No; There is no squared term!
When setting equation such as y= (3x+2)(x-5) equal to zero, what are you finding?
x-intercepts
zeros
Find the vertex for y=(x-7)(2x+6)
(2,-50)
Name the x-intercepts y=5x^2 - 10x
(0,0) and (2,0)?
Name the vertex x^2 - 4x - 5 = 0
(2,-9)
Name the x-intercept(s) for y=(12+4x)(-x+5)
(-3,0) and (5,0)
(0,-11)
Describe how the parabola would shift and change from y=x^2 to y = - 1/2 (x + 7)^2 - 4
Flips
Gets wider
Shifts 7 left and 4 down
Name the x-intercepts. y= (x-1)^2 + 1
None; the parabola has opens upward and has its vertex at (1,1)
How will f(x)=(x+7)^2 + 3 change if you shift the parabola 12 units down?
f(x) = (x+7)^2 - 9
True or False?
The coordinates of the vertex for y=x^2 -6x are at (2,-8)
Justify!
False;
(2,-8) is one of the points on the parabola but not the vertex. y=x^2-6x = x(x-6) which means the x-intercepts are at (0,0) and (6,0) so halfway between 0 and 6 is 3!