Iceycold12
Junior Member
- Joined
- Feb 24, 2012
- Messages
- 55
Hello.
What is the range of the function? 4x^2-5x+2
I have these answer choices:
A. y>=2
B. y>=5/16
C. y>=7/16
D.y>=5/8
My teacher didn't teach this very well so I'm trying to work this out from stuff I have found online. The parabola opens upward so all y values y>=k. The y coordinate value of the vertex is the range of the function, so let's find the vertex.
I. Finding x (axis of symmetry): 5/2(4) = 5/8 = 0.625
II. Plug in x to find y coordinate of the vertex
III. y=4(.625)^2-5(.625)+2
IV. y=1.5625-3.12500+2
V. y=0.4375 = 7/16
Is the Answer C. y>=7/16?
Thanks a lot.
What is the range of the function? 4x^2-5x+2
I have these answer choices:
A. y>=2
B. y>=5/16
C. y>=7/16
D.y>=5/8
My teacher didn't teach this very well so I'm trying to work this out from stuff I have found online. The parabola opens upward so all y values y>=k. The y coordinate value of the vertex is the range of the function, so let's find the vertex.
I. Finding x (axis of symmetry): 5/2(4) = 5/8 = 0.625
II. Plug in x to find y coordinate of the vertex
III. y=4(.625)^2-5(.625)+2
IV. y=1.5625-3.12500+2
V. y=0.4375 = 7/16
Is the Answer C. y>=7/16?
Thanks a lot.