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#1 2015-04-05 03:08:01

denis_g
Member
Registered: 2015-04-05
Posts: 1

Algebra 2: Quadratics

Find all values of p such that
2(x+4)(x-2p)
has a minimum value of -18.

Please explain your answer as thoroughly as possible, as I am very much stuck

Last edited by denis_g (2015-04-05 03:08:27)

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#2 2015-04-05 05:10:23

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra 2: Quadratics

Hi;

Easiest is to use the properties of a parabola ( ax^2+bx+c=0 ) which says the vertex is located at

Expand out your problem.

Equate coefficients.

a = 2

b = ( 8 - 4p )

c = - 16 p

The coordinates are

We set the y coordinate to -18 and solve for p.

p = 1 or p = - 5

Draw the graph to convince yourself that you have indeed found a minimum.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2015-04-05 05:29:59

Olinguito
Member
Registered: 2014-08-12
Posts: 649

Re: Algebra 2: Quadratics

Now find all p such that this is equal to the given minimum value.


Bassaricyon neblina

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