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# A quadratic function f(x) attains a max of 3 at x =1, the

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A quadratic function f(x) attains a max of 3 at x =1, the [#permalink]

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07 Dec 2009, 10:52
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 03 Feb 2012, 04:25, edited 1 time in total.
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07 Dec 2009, 13:13
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xcusemeplz2009 wrote:
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

Nice explanation sriharimurthy.

I did some calculation mistake,

My approach:

$$f(x)=ax^2+bx+c$$

Given for$$x=0$$, $$f(x)=1$$
so $$c=1$$
Given for $$x=1$$, $$f(x)=3$$
so $$a+b+c=3$$
$$a+b=2$$

now question is $$100a+10b+c=?$$
$$90a+10(a+b)+c$$
$$90a+21=?$$

a, b, and c are integers

now pick an answer choice for which a is an integer

$$90a+21=-159$$
$$a=-2$$

Last edited by swatirpr on 07 Dec 2009, 14:11, edited 1 time in total.
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07 Dec 2009, 13:32
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xcusemeplz2009 wrote:
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

Let the quadratic function be : $$f(x) = ax^2 + bx + c$$

Given : $$f(0) = 1$$

Substituting this in our equation we get : $$c = 1$$

Thus our function becomes : $$f(x) = ax^2 + bx + 1$$

Given : $$f(1) = 3$$ is the max value of 'x'.

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 $$f(10) = -159$$

Ps. I doubt this is a GMAT problem. Please mention the source.
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07 Dec 2009, 19:53
sriharimurthy wrote:
xcusemeplz2009 wrote:
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

Let the quadratic function be : $$f(x) = ax^2 + bx + c$$

Given : $$f(0) = 1$$

Substituting this in our equation we get : $$c = 1$$

Thus our function becomes : $$f(x) = ax^2 + bx + 1$$

Given : $$f(1) = 3$$ is the max value of 'x'.

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 $$f(10) = -159$$

Ps. I doubt this is a GMAT problem. Please mention the source.

wow!!
grt job
i solved lke this and got -159 ,but was not sure abt the ans

let f(x) = ax2 + bx +c

at x = 0 value of f(x) = 1 so equation becomes

1 = c

since it attains max of 3 at x =1

3 = a+b+c put c=1

a+b = 2

now differntiate f(x) we will get 2ax +b =0 since it attains max at x =1 relaion becomes 2a = -b

now u will get values of a=-2,b =4 nd c =1

f(x) = -2x2 + 4x + 1

now put x = 10 nd answer is -159
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08 Dec 2009, 14:39
xcusemeplz2009 wrote:
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
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11 Dec 2009, 02:32
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GMATBLACKBELT wrote:
xcusemeplz2009 wrote:
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 $$b=4$$.

$$f(x)=-2x^2+4x+1$$ --> $$f(10)=-200+40+1=-159$$

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11 Dec 2009, 15:32
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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.

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

Then, just plug in 10 for x and solve.
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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?

Posted from my mobile device
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19 Dec 2009, 10:00
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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.

$$f_{max}(1)=3=ax^2+bx+1=a*1^2+b*1+1=a+b+1$$, --> $$a+b+1=3$$ --> $$a+b=2$$.

Also the property of quadratic function:

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: $$f(x)=-2x^2+4x+1$$

Hope it's clear.
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12 Oct 2010, 22:10
Is it REALLY gmatprep's task? I've never seen this in gmatprep. Or the "tag" is wrong?
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18 Aug 2015, 18:51
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11 Sep 2015, 22:03
Bunuel wrote:
GMATBLACKBELT wrote:
xcusemeplz2009 wrote:
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 $$b=4$$.

$$f(x)=-2x^2+4x+1$$ --> $$f(10)=-200+40+1=-159$$

how should i understand that 1 is the vertex of parabola. is it the reason that x=0 ?please clarify this point
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