Finding equation with zeroes and max value

[Drake M]

[Note: Thread moved to this forum by a mentor]

Hi all, the full question is:
A parabola has zeroes at 5/2 and -3/2 and has an optimal value of 4. Determine equation of the parabola in factored form.

So I started out with 4 as my y-value on my vertex then took the zeroes(5/2 and -3/2), added them and divided by 2 to get the x-axis for the x-value in the vertex.
Subbed in (.5,4) into factored form with the zeroes,
y=a(x-r)(x-s)
4=a(.5-5/2)(.5-(-3/2)
4=a(-2)(2)
4/-4=a
-1=a
So the equation should be y=-1(x-5/2)(x-(-3/2) or y=-1(2x-5)(2x+3)
The answer sheet says its y=-1/4(2x-5)(2x+3)
I have similar numbers but don't know where I went wrong. Any help is appreciated

 

[fresh_42]

Hi all, the full question is:
A parabola has zeroes at 5/2 and -3/2 and has an optimal value of 4. Determine equation of the parabola in factored form.

So I started out with 4 as my y-value on my vertex then took the zeroes(5/2 and -3/2), added them and divided by 2 to get the x-axis for the x-value in the vertex.
Subbed in (.5,4) into factored form with the zeroes,
y=a(x-r)(x-s)
4=a(.5-5/2)(.5-(-3/2)
4=a(-2)(2)
4/-4=a
-1=a
So the equation should be y=-1(x-5/2)(x-(-3/2) or y=-1(2x-5)(2x+3)
The answer sheet says its y=-1/4(2x-5)(2x+3)
I have similar numbers but don't know where I went wrong. Any help is appreciated

How did your second equation derive? What does an optimal value of 4 really mean?

 

[Drake M]

isnt optimal value max value of y which would be the vertex

 

[SammyS]

isn't optimal value max value of y which would be the vertex

I suppose so, which would follow from 4 being the maximum value. The other common possibility being that it's a minimum, but then of course that would imply that there are no x-intercepts.

 

[fresh_42]

isnt optimal value max value of y which would be the vertex

It may be a maximum, or a minimum. Anyway, how do you usually find it and where? All you know is the y value of 4 I guess.

Where does this line come from?

4=a(.5-5/2)(.5-(-3/2)

But maybe you are using another method than me. However, to find your deviation from the given result I need to know better what you are trying to do.

EDIT: Ok, I got it now. Everything is fine but the way you got rid of the factor 1/2 in the parenthesis. You multiplied twice by two and didn't correct it outside.

 

[Drake M]

Thanks for the help guys, a buddy of mine solved it and showed me how. I wasnt simplifying x-5/2 into 2x-5 and x--3/2 into 2x+5 before is subbed in my vertex values and that's where I went wrong

 

[SammyS]

So the equation should be y=-1(x-5/2)(x-(-3/2) or y=-1(2x-5)(2x+3)
The answer sheet says its y=-1/4(2x-5)(2x+3)
I have similar numbers but don't know where I went wrong. Any help is appreciated

Those answers are equivalent. except for the part crossed out