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If 'x' is maximum at x = 1, then \(\frac{d}{dx}f(x) = 0\) at x = 1

\(\frac{d}{dx}f(x) = 2ax + b = 0\) Therefore, putting value of x =1 we get \(b = -2a\)

Now going back to \(f(1) = 3\), we get \(a + b + 1 = 3\) --> \(a - 2a + 1 = 3\)

Thus, \(a = -2\) and \(b = 4\)

So our quadratic function is : \(f(x) = -2x^2 + 4x + 1\)

Thus \(f(10) = -159\)

Answer : A

Ps. I doubt this is a GMAT problem. Please mention the source.
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A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159, b)-110, c)-180, d)-105, e)-119

Pls explain

Not sure this would be on the GMAT, seems you should use calculus for this.

ax^2+bx+c=0

f(1)=3=a+b+c f(0)=1=c

f'(x)= 2xa+b.

since the max is 3, f'(1)=0. Thus, 0=2a+b--> b=-2a --> 2=a-2a --> -2=a

f(10) = -2*100+10(4)+1 --> -159

This can be solved in the following way too:

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

\(f(0)=c=1\) --> \(f(x)=ax^2+bx+1\)

We are told that \(f_{max}(1)=a+b+1=3\), --> \(a+b=2\).

\(f_{max}\) is vertex of parabola and the \(x\) coordinate of vertex is \(-\frac{b}{2a}=1\) --> \(b=-2a\) --> \(a+b=a-2a=-a=2\) --> \(a=-2\) and \(b=4\).

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

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?

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]

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20 Oct 2017, 13:08

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