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

Question

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

Bunuel · Accepted Answer

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

Answer: A.

sriharimurthy · Answer

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

Answer : A

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

swatirpr · Answer

Nice explanation sriharimurthy.

I did some calculation mistake, 
answer is A, -159.

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

so A is the answer.

xcusemeplz2009 · Answer

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

GMATBLACKBELT · Answer

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

umangpatel03 · Answer

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

Then, just plug in 10 for x and solve.

CoIIegeGrad09 · Answer

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

Bunuel · Answer

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.

rsrighosh · Answer

I was not aware about the concept of maximum value of quadratic equation is ( \frac{-b}{2a}, c-\frac{b^2}{4a} ). This goes in my notebook.

However, I used POE technique. f(x) = ax^2 +bx+c x=1, a+b+c =3 x=0, c=1 Hence, a+b = 2 or b = 2-a So, f(10) = 100a+10b+c = 100a + 10(2-a) +c = 90a + 21 = 3(30a + 7) So, we understand the value is a multiple of 3. Hence, shortlisting a)-159, c)-180, d)-105 Now, 90a+21. the last digit can end up in 1 if a is =+ve or 9 if a is -ve Only option with last digit 9 is a) -159 Hence A.

KarishmaB · Answer

Here, the function takes the max value of 3 at x = 1. So it is a downward open parabola.

Say it is  f(x) = ax^2 + bx + c 
Since its value is 1 at x = 0, we get c = 1.

Also at x = 1,  a + b + 1 = 3 
So a + b = 2

Since the graph takes the max value at x = 1, we know that -b/2a = 1. So b = -2a.
Using this in the equation above, we get a = -2 and b = 4

So f(x) becomes  -2x^2 + 4x + 1 .

When x = 10, we get -159.

Answer (A)

smmc29 · Answer

I used this form:

y = +- a(x-k)^2 + c

Since the question specifically indicates that a maximum value is present, then the parabola should be facing downwards (a should be negative)

y = -a(x-k)^2 + c

Knowing that the function is maximum at x = 1, therefore, k = 1. We can see here that whatever happens, the term (x-k)^2 is greater than or equal to zero. In order to maximize, we must make sure that this part of the expression will be minimum; hence, equal to 0.

we now know that k = 1, and c = 3.

y = -a(x-1)^2 + 3

To get the value of a, use the second constraint. At x = 0, a should be equal to 2 so that y = 1.

y = -2(x-1)^2 + 3

Substitute x = 10 to get the answer.

Answer is -159.

Fdambro294 · Answer

When a quadratic function opens downward, the apex or vertex is the maximum value of f(x) that can be achieved.

The quadratic function for a downward opening parabola should take the form:

F(X) = a(X)^2 + b(X) + c

Where the value of the a coefficient will be Negative

The maximum value or vertex will occur at the vertical line given by the X -Coordinate of:

X = -b / (2a)

Since we are told that this maximum occurs at f(x) = 3 when X = 1

1 = -b / (2a)

2a = -b

b = -2a

Additionally, we are told that we X = 0, the Y -Intercept will be 1 ————> c = 1

Plugging in:
b = -2a
c = 1

f(x) = a(X)^2 - 2a(X) + 1

Finally, we can plug in the coordinate of (1 , 3) to find the value of a

3 = a(1)^2 - 2a(1) + 1

2 = a - 2a

2 = -a

Or

a = -2

And

b = +4

The function will take the form:

F(X) = -2(X)^2 + 4(X) + 1

Plugging in the X coordinate of 10, we get:

F(10) = -200 + 40 + 1

Answer

-159

(A)

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NirupaD · Answer

This solution is easier using co-ordinate geometry, rather than differencian formula f(max)=-b/2a

lnyngayan · Answer

Could you explain more on this?

f_{max}  is vertex of parabola and the  x  coordinate of vertex is  -\frac{b}{2a}=1

Many thanks!

Bunuel · Answer

Parabola

A parabola is the graph associated with a quadratic function, i.e. a function of the form  y=ax^2+bx+c .

https://gmatclub.com/forum/download/file.php?id=10339

The general or standard form of a quadratic function is  y =ax^2+bx+c , or in function form,  f(x)=ax^2+bx+c , where  x  is the independent variable,  y  is the dependent variable, and  a ,  b , and  c  are constants.

The larger the absolute value of  a , the steeper (or thinner) the parabola is, since the value of y is increased more quickly.
 
  If  a  is positive, the parabola opens upward, if negative, the parabola opens downward.

x-intercepts:  The x-intercepts,  if any , are also called the roots of the function. The x-intercepts are the solutions to the equation  0=ax^2+bx+c  and can be calculated by the formula:
 x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}  and  x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}

Expression  b^2-4ac  is called  discriminant :
   If discriminant is positive parabola has two intercepts with x-axis; 
  If discriminant is negative parabola has no intercepts with x-axis; 
  If discriminant is zero parabola has one intercept with x-axis (tangent point).

y-intercept:  Given  y =ax^2+bx+c , the y-intercept is  c , as y intercept means the value of y when x=0.

Vertex:  The vertex represents the maximum (or minimum) value of the function, and is very important in calculus.

The vertex of the parabola is located at point  (-\frac{b}{2a},   c-\frac{b^2}{4a}) .
Note: typically just  -\frac{b}{2a},  is calculated and plugged in for x to find y.

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Regor60 · Answer

The fewer formulas to memorize, the better.

You simply need to remember that a parabola is symmetrical about its vertex.

So if the vertex is at X=1 as is stated, then the same values for Y will obtain for equal X distance from X=1.

So since it's given that Y=1 at X=0, Y will also be equal to 1 at X=2.

This will give you enough information to solve for the coefficients of X^2 and X.

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

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