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Re: function problem [#permalink]
11 Dec 2009, 15:32

You've got 3 points and 3 equations, so you could use a system of equations to solve. Coordinates: (1,3) Given, (0,1) Given, and (2,1) Implied. I got (2,1) by doing a really rough sketch and seeing that (2,1) is a mirror of (0,1) with (1,3) as the max.

3 systems in quadratic form: a(1)^2 + b(1) + c = 3 a(0)^2 + b(0) + c = 1 a(2)^2 + b(2) + c = 1

a + b + c = 3 c = 1 4a + 2b + c = 1

Solving this would yield a = -2, b = 4, c = 1. f(x) = -2x^2 + 4x + 1

Re: function problem [#permalink]
19 Dec 2009, 08:33

Hey guys,

I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?

Re: function problem [#permalink]
19 Dec 2009, 10:00

Expert's post

CoIIegeGrad09 wrote:

Hey guys,

I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?

Posted from my mobile device

We have the function f(x)=ax^2+bx+c.

Stem: the value of the function at x =0 is 1.

Substitute x by 0 --> f(0)=1=a*0^2+b*0+c --> c=1, hence the function becomes: f(x)=ax^2+bx+1.

Stem: A quadratic function f(x) attains a max of 3 at x =1.

The maximum (or minimum if a>0) of the quadratic function f(x)=ax^2+bx+c is the y coordinate (the value of f(x)) of the vertex of the given quadratic function, which is parabola.

The vertex of the parabola is located at the point (-\frac{b}{2a},c-\frac{b^2}{4a}).

Given: A quadratic function f(x) attains a max of 3 at x =1 --> coordinates of the vertex are (1, 3) --> -\frac{b}{2a}=1, c-\frac{b^2}{4a}=3

So we have: a+b=2 and: -\frac{b}{2a}=1

From this a=-2 and b=4. so the function is: f(x)=-2x^2+4x+1

Re: A quadratic function f(x) attains a max of 3 at x =1, the [#permalink]
07 Jul 2014, 00:40

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