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December 20, 2014

December 20, 2014

Posted by **Sarskia** on Sunday, August 22, 2010 at 9:51pm.

y=3x-3/4x^2

Use calculus to find the maximum height of the curved parabolic section.

How do I do this step by step?

- Calculus -
**MathMate**, Sunday, August 22, 2010 at 10:03pmIf 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)

- Calculus -
**Reiny**, Sunday, August 22, 2010 at 10:05pmyour 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 ....

- Calculus -
**Reiny**, Sunday, August 22, 2010 at 10:06pmgo with Damon

I skimmed over the "parabolic section" of your post.

- Calculus -
**Sarskia**, Sunday, August 22, 2010 at 10:09pm..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.

- Calculus -
**MathMate**, Sunday, August 22, 2010 at 10:16pmYou 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

- Calculus -
**Sarskia**, Sunday, August 22, 2010 at 10:18pmOkay, thank you very much!

- Calculus -
**MathMate**, Sunday, August 22, 2010 at 11:20pmYou're welcome!

- Calculus -
**Sarskia**, Monday, August 23, 2010 at 4:13amI had a run over it myself and got the maxima as (-2,-3) same numbers, but negative form?

- Calculus -
**MathMate**, Monday, August 23, 2010 at 8:57amCheck 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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