Posted by Sarskia on Sunday, August 22, 2010 at 9:51pm.
If you have done functions before calculus as most students do, you can rewrite the equation in the standard form:
f(x)=a(x-h)+k
where the vertex is found at
(h,k)
or
f(x)=-(3/4)(x-2)²+3
where the vertex is found at (2,3).
Confirm by Calculus:
f(x)=3x-3/4x^2
f'(x)=3-(3/2)x
At the vertex, f'(x)=0
x=3*(2/3)=2
y=3(2)-(3/4)(2²)
=6-3
=3
So the vertex is at (2,3)
your equation is ambiguous
do you mean
y = (3x-3)/(4x^2) ? or
y = 3x - 3/(4x^2) ? or
y = ((3x-3)/4)(x^2) or
y = 3x - (3/4)x^2 ? or ....
go with Damon
I skimmed over the "parabolic section" of your post.
..The last one.
Confirm by Calculus:
f(x)=3x-3/4x^2
f'(x)=3-(3/2)x
At the vertex, f'(x)=0
x=3*(2/3)=2
y=3(2)-(3/4)(2²)
=6-3
=3
So the vertex is at (2,3)
A question regarding the first step (differentiation) How do you differentiate the fractions?
I'm stuck.. And clearly thick. Haha.
You do not really differentiate the fractional numbers, they are just the coefficients which just "stick around".
f(x)=ax²+bx
f'(x)=2ax+b
So in the above,
a=-(3/4), b=3
f'(x)=2(-3/4)x+3
=-(3/2)x+3
Okay, thank you very much!
You're welcome!
I had a run over it myself and got the maxima as (-2,-3) same numbers, but negative form?
Check your work, draw a graphical sketch to convince yourself that there is only one maximum.
Also, f(2)=3, and f(-2)=-9, so (-2,-3) does not lie on the curve.
If all fails, post your working for a check.
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