What is the range of the function y=4x^2-5x+2

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.
 
The y coordinate value of the vertex is the range

I think that you understand what you were trying to say, but the above is a misstatement.

For parabolas that open upward, the y-coordinate of the vertex is only the lowest value in the range (not the range itself).


5/2(4) = 5/8

Again, you understand the arithmetic, but you mistyped the grouping symbols above. Five-halves times four is not 5/8.

5/(2*4) = 5/8

Aside from these minor issues, you did very good with on-line self-study. :cool:
 
For parabolas that open upward, the y-coordinate of the vertex is only the lowest value in the range (not the range itself)

And a parabola that opens downward the y-coordinate of the vertex is the max value and the equation would be y=<k?
 
Top