Sum of coefficients

By method of differences,as $f(x)$ is a quadratic polynomial, the first difference line should be a arithmetic progression. Thus $f(2)-f(0)=1,f(8)-f(6)=4$ . And finally $f(8)=f(0)+[f(2)-f(0)]+[f(4)-f(2)]+[f(6)-f(4)]+[f(8)-f(6)]=1+1+2+3+4=11$

We can assume $f(x)$ takes on the form $ax^2 + bx + c$ since it is quadratic. Since $f(0) = 1$ , we can deduce that $c = 1$ . By substituting the values 2, 4, and 6 we get:

$f(4) - f(2) = 16a + 4b + 1 - (4a + 2b + 1) = 12a + 2b = 2$ $f(6) - f(4) = 36a + 6b + 1 - (16a + 4b + 1) = 20a + 2b = 3$

Given these two equations we can solve for a and b, which are $\frac{1}{8}$ and $\frac{1}{4}$ , respectively. Plugging these two values into the original quadratic expression, we arrive at $f(x) = \frac{1}{8}x^2 + \frac{1}{4}x + 1$ . Thus $f(8) = 8 + 2 + 1 = 11$ .

General form for quadratic function:

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

Using the first piece of given information:

$f(0) = a (0 ^ {2}) + b (0) + c = 1$

$c=1$

Using second piece of given information:

$f(4) - f(2) = a(4 ^ {2}) + b (4) + 1 - (a (2 ^ {2}) + b (2) + 1)$

$f(4) - f(2) = 12 a + 2 b =2$

Using third piece of given information:

$f(6) -f(4) = a(6 ^ {2}) + b (6) + 1 - (a (4 ^ {2}) + b (4) + 1)$

$f(6) -f(4) = 20 a + 2 b =3$

Combine second and third parts:

$12 a + 2 b =2$

$20 a + 2 b =3$

$-8a = -1$

$a = \frac{1}{8}$

Substitute into one of the equations to find $b$ :

$12 \frac{1}{8} + 2 b =2$

$b = \frac{1}{4}$

Substitute $a = \frac{1}{8}, b = \frac{1}{4}, c=1$ into $f(x) = ax ^ {2} + bx + c$ :

$f(x) = \frac{1}{8}x ^ {2} + \frac{1}{4}x + 1$

$f(8) = \frac{1}{8}(8) ^ {2} + \frac{1}{4}(8) + 1$

$f(8) = 11$

Let $g : \mathbb{R} \to \mathbb{R}$ be such that $g(x) = f(2+x) - f(x)$ . Since $f$ is a quadratic, $g$ (which is a difference function of $f$ ) is a linear. We are given $g(2) = 2$ and $g(4) = 3$ , and so $g(0) = 1$ and $g(6) = 4$ . This give us $f(8) - f(0) = g(6) + g(4) + g(2) + g(0) = 10$ , and so $f(8) = 11$ .

For a quadratic, $\frac{f(8)-f(0)}{8}=\frac{f(6)-f(2)}{4}=f'(4)$ . (The difference quotient is the derivative at the midpoint.) Adding two of the given equations, we find that $f(6)-f(2)=5$ . Thus $\frac{f(8)-1}{8}=\frac{5}{4}$ and $f(8)=\boxed{11.}$

After looking over the Method of Differences , and using $2n$ instead of $n$ , we can create, initially, the following table

$\begin{array}{l|lll} 2n & f(2n) & D_1(2n) & D_2(2n) \\ \hline 0 & 1 & ~ & ~ \\ 2 & ~ & 2 & ~ \\ 4 & ~ & 3 & ~ \\ 6 & ~ & ~ & ~ \\ \end{array}$

Recognize that, since we are dealing with a quadratic, $D_2$ is constant. Thus, the following can be deduced

$\begin{array}{l|lll} 2n & f(2n) & D_1(2n) & D_2(2n) \\ \hline 0 & 1 & 1 & 1 \\ 2 & 2 & 2 & 1 \\ 4 & 4 & 3 & 1 \\ 6 & 7 & 4 & 1 \\ \end{array}$

Continuing to work backwards in the table, one can find $f(8)=f(6)+D_1(6)=7+4=11$ .

$f(0) = a*0^2 + b*0 + c$

$\Rightarrow c = 1$

$f(4) - f(2) = a*4^2 + b*4 + c - a*2^2 - b*2 - c$

$\Rightarrow$ 2 = 12a + 2b

$f(6) - f(4) = a*6^2 + b*6 + c - a*4^2 - b*4 - c$

$\Rightarrow$ 3 = 20a + 2b

$3 - 2 = 20a + 2b - 12a - 2b$ $\Rightarrow a = \frac{1}{8}$

$b = 1 - 6*a$ $\Rightarrow b = 1 - 6*\frac{1}{8}$ $\Rightarrow b = \frac{1}{4}$

.

$f(8) = a*8^2 + b*8 + c$

$\Rightarrow f(8) = \frac{1}{8}*8^2 + \frac{1}{4}*8 + 1$

$\Rightarrow f(8) = 8+2+1 = \boxed{11}$

$f(x)$ is a quadratic polynomial, so we can write it as:

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

Subsituting this in the equations gives:

1. $0a + 0b + c = 1 \Leftrightarrow c = 1$

2. $16a + 4b + 1 - 4a - 2b - 1 = 12a + 2b = 2$

3. $36a + 6b + 1 - 16a - 4b - 1 = 20a + 2b = 3$

Solving 2 and 3 for $a$ and $b$ :

$2b = 2 - 12a \Rightarrow 20a - 2 + 12a = 32a - 2 = 3 \Leftrightarrow a = \frac{1}{8}$

$\frac{12}{8} - 2 = 2b \Leftrightarrow b = \frac{1}{4}$

Now we that know all the coefficients, we can simply calculate $f(8)$ :

$\frac{1}{8} \times 8^2 + \frac{1}{4} \times 8 + 1 = 11$

If $f(x)$ is a quadratic polynomial, then it must be in the form $ax^2+bx+c$ . Plugging in $x=0$ yields $f(0)=a \cdot 0^2+b \cdot 0+c=0+0+c=c$ . Since we know that $f(0)=1$ and $f(0)=c$ , then $f(0)=c=1 \Rightarrow c=1$ . Therefore, we know that $f(x)$ is in the form $ax^2+bx+1$ . When plugging in $f(2)$ , $f(4)$ , and $f(6)$ , we have $f(2)=4a+2b+1$ , $f(4)=16a+4b+1$ , and $f(6)=36a+6b+1$ . Therefore, $f(4)-f(2)=16a+4b+1-(4a+2b+1)=16a+4b+1-4a-2b-1=(16a-4a)+(4b-2b)+(1-1)=12a+2b$ . This is equal to 2, by the given condition $f(4)-f(2)=2$ . Also, $f(6)-f(4)=(36a+6b+1)-(16a+4b+1)=(36a-16a)+(6b-4b)+(1-1)=20a+2b$ . This is equal to 3, by the given condition. Here, we have a system of equations: $12a+2b=2$ and $20a+2b=3$ . Subtracting the second equation from the first, we have $8a=1 \Rightarrow a=\frac{1}{8}$ . Plugging this in back to the first equation from the system of equations, we have $12(\frac{1}{8})+2b=2 \Rightarrow \frac{3}{2}+2b=2 \Rightarrow 2b=\frac{1}{2} \Rightarrow b=\frac{1}{4}.$ . Now, we have our complete polynomial: $f(x)=\frac{1}{8} x^2+\frac{1}{4}x+1$ , and $f(8)=\frac{1}{8} \cdot 8^2+\frac{1}{4} \cdot 8+1=\frac{1}{8} \cdot 64+\frac{1}{4} \cdot 8+1=8+2+1=\boxed{11}$

Construct $f(x)=ax^2+bx+c$ , we get $f(0)=c=1$

Substituting $x$ with $2,4$ and $6$ , we get

$f(4)-f(2)=12a+2b=2 \ldots (1)$

$f(6)-f(4)=20a+2b=3 \ldots (2)$

from $(1)$ and $(2)$ , we can easily get $a=\frac {1}{8}$ and $b=\frac {1}{4}$ . Thus $f(8)=\frac {1}{8} \cdot 8^2+\frac {1}{4} \cdot 8+1=11$

Denote the polynomial $P(x)$ by $ax^2+bx+c$ , we have:

$\begin{matrix} P(0) = c \\ P(2) = 4a+2b+c \\ P(4) = 16a+4b+c \\ P(6) = 36a+6b+c \end{matrix}$

According to the problem, we have:

$\left\{ \begin{matrix} c=1 \\ 12a+2b = 2 \\ 20a+2b = 3 \end{matrix} \right.$

Hence we obtain:

$\begin{matrix} a=\frac18 \\ b=\frac 14 \\ c=1 \end{matrix}$

So $P(x) = \frac {x^2}{8} + \frac {x}{4} +1$ , therefore $P(8) = 11$ .

Since $f(x)$ is a quadratic, it can be written in the form $ax^2+bx+c$ .

$f(0)=1$ , so $a(0)^2+b(0)+c=1$ , or $c=1$ . Therefore, $f(x)=ax^2+bx+1$ .

$f(4)-f(2)=2$ , so $16a+4b+1-(4a+2b+1)=2$ , or $12a+2b=2$ .

$f(6)-f(4)=3$ , so $36a+6b+1-(16a+4b+1)=3$ , or $20a+2b=3$ .

Therefore, $f(8)=64a+8b+1=2((12a+2b)+(20a+2b))+1$ .

Substituting in the values for $12a+2b$ and $20a+2b$ which we got earlier, we get that $f(8)=2(3)+2(2)+1=\fbox{11}$ .

We have a quadratic polynimial represented by ax^2 + bx + c If f(0) equals to 1 this means that a 0^2 + b 0 + c = 1 And so, we get that c = 1 Once we know that we can easily find the other values using a sistem of equations. We know from f(4) - f(2) = 0 that 16a + 4b + 1 - (4a + 2b + 1) = 2, which simplified is 6a + b = 1 From f(6) - f(4) we get this other equation: 20a + 2b = 3. * Solving this two linear equations we get that * a = 1/8 and b = 1/4. * Now we can assume that our polynomial has this estructure: * 1/8x^2 + 1/4x + 1= 0 And now we can calculate f(8) which equals to 11 . SOLVED

We know that quadratic equation have maximum degree of 2.

So we write the general equation for quadratic equation is :

ax^2 + bx + c = 0

We have f(0) = 1, so 0 + 0 + c = 1

c = 1 ... (1)

f(4)-f(2) = 2. so

16a + 4b + 1 - 4a - 2b - 1 = 2

12a + 2b = 2 ... (2)

f(6)-f(4) = 3, so

36a + 6b + 1 -16a -4b -1 = 3

20a + 2b = 3 ... (3)

Solving equation (2) and (3) we have :

a = 1/8 and b = 1/4

Substitute that to f(8) = 64a + 8b + 1 = 64(1/8) + 8(1/4) + 1 = 11

First, note that, since $f(x)$ is quadratic and $f(0) = 1$ , we can say that it is of the form $f(x) = ax^2 + bx + 1$ . Now, we are given two conditions to solve for $a$ and $b$ . From the second equation, we have $(16a + 4b + 1) - (4a + 2b + 1) = 2 \implies 12a + 2b = 2.$ From the second equation, we have $(36a + 6b + 1) - (16a + 4b + 1) = 3 \implies 20a + 2b = 3.$ Note that we want $f(8) = 64a + 8b + 1 = 4 \cdot (16a + 2b) + 1.$ Referring to our previous equations, we have $f(8) = 4 \cdot (12a + 2b + 4a) + 1 = 4 \cdot (2 + 4a) + 1 = 9 + 16a.$ Now, subtracting our original two equations, we have $8a = 1$ , so $16a = 2$ and $f(8) = 9 + 16a = 9 + 2 = \boxed {11}$ .

let the quadratic equation be in the form ax^2 + bx + c

putting the values and soling by taking f(0,2,3) respectively we get a = 1/8 b= 1/4 c = 1

now putting the value as f(8)= 64a+8b+c = 64 X 1/8 + 8 X 1/4 + 1 = 11

Let f(x) = a $x^{2}$ +bx+c

f(0) = c = 1

f(4)-f(2) = 16a+4b+c-4a-2b-c = 2

12a+2b = 2 -------------------(1)

6a+b = 1

f(6)-f(4) = 36a+6b+c-16a-4b-c = 3

20a+2b = 3 -------------------(2)

(2)-(1): 8a = 1

a = $\frac{1}{8}$

b = 1- $\frac{6}{8}$ = $\frac{1}{4}$

c = 1

f(x) = $\frac{1}{8}$ $x^{2}$ + $\frac{1}{4}$ x+1

Therefore, when x=8, f(8) = 11

f(x) = ax" + bx + c f(0) = 1 c = 1

f(4) - f(2) = 2 16a + 4b + 1 - (4a + 2b + 1) = 2 12a + 2b = 2 6a + b = 1 ... (1)

f(6) - f(4) = 3 36a + 6b + 1 - (16a + 4b + 1) = 3 20a + 2b = 3 10a + b = 3/2 ... (2)

eliminate (1) & (2) 4a = 1/2 a = 1/8 .. (3)

6a + b = 1 [subs (3)] 6/8 + b = 1 b = 2/8 = 1/4

So, f(8) = (1/8) . 8" + (1/4) . 8 + 1 = 8 + 2 +1 = 11

Since it is the question of comparing Coefficients. So we have: f(0) = 1, f(4) - f(2) = 2 and f(6) - f(4) = 3. therefore, f(4) - f(2) = f(6) - f(4) = f(2) since, [f(4) - f(2)] + [f(4) - f(2)] + [f(6) - f(4)] + [f(6) - f(4)] + [f(0)] f(8) = 2 + 2 + 3 + 3 + 1 = 11. ans

Now , the question will arise in your mind that why would we add f(0) the answer is f(0) is given in the question and it is the one of the coefficient that's why we add f(0).

Just see , if quadratic equation is , $ax^2+bx+c$ then since $f(0)=1 \implies c=1$
The other two relation give $6a+b=1 , 20a+2b=3$ which on solving gives $a=1/8 , b = 1/4$ thus the result :)

Let f(x)=ax^2+bx+c => f(0) =1 ; change x into f(x) with f(2);f(4) and f(6), we have a=1/8; b=1/4 => f(8)=8.(1/8)^2 + 8.(1/4) + 1 =11

Let $f(x) = Ax^{2} + Bx + C$ Then $f(0) = 1 \implies C = 1$

$\therefore f(x) = Ax^{2} + Bx + 1$

Now, $f(2x) - f(2x-2) = 8Ax - 4A + 2B$

putting $x = 2$ and $x = 3$ successively; we get :

$f(4) - f(2) = 12A + 2B = 2$

$f(6) - f(4) = 20A + 2B = 3$

On solving the systems of equations we get $A = \frac {1}{8}$ and $B = \frac {1}{4}$

So, $f(x) = \frac {x^{2}}{8} + \frac {x}{4} + 1 \therefore f(8) = 11$

Each quadratic polynomial can be written in the form of $ax^2+bx+c$ , where $a,b$ and $c$ are the real coefficients of the polynomial.

Having that in mind, we may use the first equation to find the value of $c$ in our polynomial

$f(0)=a\cdot0+b\cdot0+c \Rightarrow a\cdot0+b\cdot0+c=1 \Rightarrow c=1$ .

Having the value of $c$ in mind, we may now create a system using the following $2$ equations, since $f(4)=16a+4b+1, f(2)=4a+2b+1$ and $f(6)=36a+6b+1$ . Therefore,

$\left\{ \begin{array}{l l} 12a+2b=2& \quad \\ 20a+2b=3 & \quad \end{array} \right.$

from which we can easily find that $a=\frac{1}{8}$ and $b=\frac{1}{4}$ .

The polynomial we are looking for is $\frac{1}{8}x^2+\frac{1}{4}x+1$ . With that in mind, we can easily see that $f(8)=\frac{1}{8}64+\frac{1}{4}8+1 \Rightarrow f(8)=8+2+1 \Rightarrow f(8)=11$ , which is the correct answer to the problem.

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

$f(0) = 1 \implies c = 1$

$f(x) = ax^2 + bx + 1$

$f(4)-f(2) = 16a + 4b - 4a - 2b = 2$

$f(4)-f(2) = 36a + 6b - 16a - 4b = 3$

\­(f(4)-f(2) = 12a + 2b = 2)

\­(f(4)-f(2) = 20a + 2b = 3 )

Therefore,

\­(8a= 1 \implies a = \frac{1}{8} )

Hence,

$\frac{12}{8} + 2b = 2 \implies b = \frac{1}{4}$

Finally,

\­(f(x) = \frac{1}{8}x^2 + \frac{1}{4}x + 1 )

So $f(8)=8 + 2 + 1=11$

Let f(x)=ax^2 + bx +c since f(0)=1, we can conclude that c=1, now we have, f(x)=ax^2 + bx +1 since f(4)-f(2)=2 16a+4b+1-(4a+2b+1)=2 6a+b=1................equation 1

also, f(6)-f(4)=3 36a+6b+1-(16a+4b+1)=3 20a+2b=3..............................equation 2

solve a and b using equation 1 and 2, we get a=1/8 b=1/4

finally, f(x)=(1/8)x^2 + (1/4)x +1

and f(8)=(1/8) 8^2 + (1/4) 8 +1 f(8)=11

For quadratic polynomial use ax^2+bx+c=0. Find out f(0),f(2),f(4),f(6)... and put them in given equations to find a,b &c.. finally get your f(x) and put x=8.!!

Let f(x) = ax^{2} + bx + c

f(0) = 1, c = 1.

f(2) = 4a + 2b + 1 , f(4) = 16a + 4b +1 , f(6) = 36a + 6b + 1.

f(4) - f(2) = 2

12a + 2b = 2 ...........(i)

f(6) - f(4) = 3

20a + 2b = 3 ............(ii)

On solving (i) & (ii), a = 1/8, b = 1/4.

Therefore, f(x) = (1/8)x^{2} + (1/4)x + 1, and f(8) = 8 + 2 +1 = 11 (Ans.)

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

$f(0)=1$

$a(0)^2+b(0)+c=1$

$c=1$

$f(4)-f(2)=2$

$[a(4)^2+b(4)+c]-[a(2)^2+b(2)+c]=2$

$12a+2b=2-----(1)$

$f(6)-f(4)=3$

$[a(6)^2+b(6)+c]-[a(4)^2+b(4)+c]=3$

$20a+2b=3-----(2)$

From $(1)$ and $(2)$ , we have

$a=\frac{1}{8}$

$b=\frac{1}{4}$

Now, we arrange them and get

$f(x)=\frac{1}{8}x^2+\frac{1}{4}x+1$

So, $f(8)=11$

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

f(0) = 1, c = 1

f(4)-f(2) = 2, 16a+4b+1-(4a+2b+1) = 2, 12a+2b=2

f(6)-f(4) = 3, 36a+6b+1-(16a+4b+1) = 3, 20a+2b=3

a= 1/8, b=1/4, f(8)=11

By method of differences, $1 = f(2) - f(0) = f(2) - 1$ so $f(2) = 2, f(4) = 6, f(6) = 7$ and $f(8) = f(6) + 4 = 7 + 4 = 11$ .

f(x) = ax^2+bx+c, já que f(x) é um polinômio quadrático.Assim, f(x) = ax^2 + bx + 1. Utilizando as equações de recorrência e somando-as chegamos em a=1/8 e b=1/4. Assim, f(x) = (1/8).x^2 + (1/4)x + 1. Para f(8) achamos 11.

Let Quadratic Equation be f(x)= ax^2 + bx + c Its given f(0)=1 so from here we get c=1 now f(4) - f(2) = 2 we have => 6a+b=1 and f(6) - f(4) = 3 => 20a+2b=3 solving for 'a' and 'b' we have => a=1/8 and b=1/4 so f(8) = (8) (8) (1/8) + (8)*(1/4) +1 = 11 :)

The polynomial is ax^2 + bx + 1. Now substitute the other two equations, which give you, 12a + 2b=2; 20a + 2b= 3. Hence 8a=1, 4b=1; hence f(8)=64a + 8b +1= 8 + 2 + 1 = 11.

Solve the 3 equations to get the value of coefficients i.e. a,b and c. c=1, 12a+2b+c=2 20a+2b+c=3

Which gives:- a = 1/8, b = 1/4, c = 1

So now for f(8)=64 1/8 + 8 1/4 +1=11

Let, the quadratic polynomial is f(x) = a $x^2$ + bx + c
f(0) = c = 1 [given]
So, f(4) - f(2) = $4^2$ a + 4b + 1 - $2^2$ a - 2b - 1 = 2 [given] ⇒12a + 2b = 2...........(1)
and, f(6) - f(4) = $6^2$ a + 6b + 1 - $4^2$ a - 4b - 1 = 3 [given]
⇒20a + 2b = 3...........(2)
By equating '2b' from both equations (1) and (2), we shall get, ⇒2 - 12a = 3 - 20a
⇒8a = 1 ⇒a = \frac {1}{8} \
By putting the value of 'a' in equation (1), we shall get,
b = \frac {1}{4}
So, f(8) = \frac {1}{8} * $8^2$ + \frac {1}{4} *8 + 1 = 8 + 2 + 1 = 11[ANSWER]

Quadratic polynomial has a form: $f(x)=ax^2+bx+c$

So by replacing values we get:

$a \times 0^2 + b \times 0 +c = 1$

$4^2a+4b+c-\left( 2^2a+2b+c \right)$

$6^2a+6b+c-\left( 4^2a+4b+c \right)$

By solving these equations we get that

$a=\frac{1}{8}$

$b=\frac{1}{4}$

$c=1$

Now we know that our function has a form: $f(x) = \frac{1}{8}x^2+\frac{1}{4}x+1$

By replacing 8 in the x we get 11

Let $f(x)=ax^2+bx+c$ . Since f(0)=1 => c=1; $f(4)-f(2)=a(4^2-2^2)+ b(4-2)=2$ ; and $f(6)-f(4)= a(6^2-4^2)+b(6-4)=3$ ;

We get $12a+2b=2$ and $20a+2b=3$ .

Solving the equations we have $a=1/8$ and $b=1/4$ .

So $f(8)=a*8^2+b*8+c=64/8+8/4+1=8+2+1=11$

Let us suppose that the quadratic polynomial is ax^2+bx+c. From the first condition, we have that c =1. Now , if we take the second condition, we get the following equation: 12a + 2b = 2 [EQUATION (A)] and if we take the third condition, we get 20a + 2b = 3 [EQUATION (B)] Our job is to find f(8), i.e. 64a + 8b +1. Now let us add equations (A) and (B). We get
32a + 4b = 5. Interestingly, we get 64a + 8b +1 = 2(32a +4b) +1 = 2(5) + 1 = 11. Hence f(8) = 11. (I hope it is clear how we got these linear equations, (by substituting for x).)

Let the quadratic be $ax^2+bx+c$ Thus we get a system of equations:

$0a+0b+c=1$

$(16a^2+4b+c)-(4a^2+2b+c)=2$

$(36a^2+6b+c)-(16a^2+4b+c)=3$

Linear combination gives us $a=\frac{1}{8}, b=\frac{1}{4}, c=1$

Thus the answer is $\frac{64}{8}+\frac{8}{4}+1=\boxed{11}$

F(0) = AX^2+BX+C=1 => C=1 F(4)-F(2)=2 => 16A+4B+1-(4A+2B+1)=2 12A+2B=2 => 6A+B=1 F(6)-F(4)=3 => 36A+6B+1-(16A+4B+1)=3 => 20A + 2B = 3 => 10A+B=3/2

Solving the system: 10A+B=3/2 6A+B=1 (-1) 4A=1/2 => A = 1/8 and B = 1/4

Hence, 1/8 * 64+1/4 * 8+1=8+2+1=11

f(x)=ax²+bx+c

f(0)=c=1

f(4)-f(2)=12a+2b=2 ...(i)

f(6)-f(4)=20a+2b=3 ...(ii)

f(8)=64a+8b+1 ...(iii)

mas, fazendo: 2.(i)+2.(ii), obtemos: 64a+8b=10

logo, substituindo em (iii), temos f(8)=10+1=11

because quadratic polynomial, so d2 is constant, from f(4)-f(2)=2 and f(6)-f(4)=3 we know that d2=1 then f(2)-f(0)=1, f(2)=2, f(4)=3, f(6)=7 and f(8)-f(6)=4, f(8)=11

c=1 (f(0)=1)

f(4) - f(2) = 2, then 16a + 4b + 1 - 4a - 2b - 1 = 2 12 a + 2b = 2 6a + b = 1 b= 1 - 6a

f(8) - f(4) = 3, then 36a + 6b + 1 - 16a - 4b - 1 = 3 20a + 2b = 3

Then 20a + 2 - 12a = 3 8a = 3-2 8a = 1 a = 1/8

b= 1 -6/8 b= 2/8 = 1/4

1/8 x 64 + 1/4 x 8 + 1 8 + 2 + 1 11

f(x) = a x^2 + b x + c => f(0) = c, so c = 1 By computing f in terms of a and b for the remaining two relationships, we get a system of two linear equations in two unknowns: 12 a + 2 b = 2 20 a + 2 b = 3. It follows a = 1/8, b = 1/4.

Let f(x) = ax^2 + bx + c then f(o) = 1 gives c = 1 f(4) - f(2) = 2 gives 12a + 2b = 2 ........(1) f(6) - f(4) = 3 gives 20a +2b = 3 ....... (2)

solving for a and b from (1) and (2) we shal get a= 1/8 and b= 2/8

Put x = 8, a = 1/8 , b= 2/8 , c = 1 in the first expression we get f(8) = 11

Assume , The Quadratic to be $f(x)=ax^2+bx+c$ . Now, Using the First Constraint we find that $c=1$ . Similarly using other two equations we found $a=\frac{1}{8}$ , $b=\frac{1}{4}$ .

So, Our Quadratic becomes $f(x)=\frac{1}{8}x^2+\frac{1}{4}x+1$ .

Then, $f(8)=11$ .

Let $f(x) = ax^2 + bx + c$ , first equality implies $c = 1$ and two other equalities give us system $$ 12a + 2b = 2$$ $$20a + 2b = 3$$ Solving this gives us $(a,b,c) = \left(\frac{1}{8},\frac{1}{4},1\right)$ So $$f(8) = \frac{8^2}{8} + \frac{8}{4} + 1 = \boxed{11}$$

Since $f(x)$ is a quadratic polynomial, let $f(x) = ax^2 + bx + c$ . Since $f(0) = 1$ , substituting $x = 0$ into $f(x)$ , $c = 1$ . Hence $f(x)$ can be rewritten as $f(x) = ax^2 + bx + 1$ .

Now, $f(4) - f(2) = 2$ . By substitution, we get $16a + 4b + 1 - (4a + 2b + 1) = 2 \Rightarrow 12a + 2b = 2$ .

Repeat this for $f(6) - f(4) = 3$ . In this case we get $36a + 6b + 1 - (16a + 4b + 1) = 3 \Rightarrow 20a + 2b = 3$ .

We now have a system of linear equations that we can solve for: $\begin{cases} 12a + 2b = 2 - (1)\\ 20a + 2b = 3 - (2) \end{cases}$

$(2) - (1): 8a = 1 \Rightarrow a = \frac{1}{8}$ . Substituting $a = \frac{1}{8}$ into $(1)$ , $\frac{12}{8} + 2b = 2 \Rightarrow b = \frac{1}{4}$ . Hence the polynomial $f(x) = \frac{1}{8}x^2 + \frac{1}{4}x + 1$ .

Finding $f(8)$ is now just a matter of substituting $x = 8$ into $f(x)$ . $f(8) = \frac{1}{8}(8^2) + \frac{1}{4}(8) + 1 = \fbox{11}$ .

let equation be ax^2+bx+c=0 f(0)=1=>c=1 f(4)−f(2)=2=> 6a+b=1 f(6)−f(4)=3=>10a+b=3/2 on solving a=1/8 b=1/4 and c=1 then quadratic equation is (1/8)x^2+(1/4)x+1=0 and f(8)=11

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

$f\left(0\right) = c = 1$

$f\left(4\right) - f\left(2\right) = \left(16a+4b+1\right) - \left(4a+2b+1\right)$

$= 12a + 2b = 2$

$b = 1-6a$

$f\left(6\right) - f\left(4\right) = \left(36a+6b+1\right) - \left(16a+4b+1\right) = 3$

$20a + 2b = 3$

$20a + 2 -12a = 3$

$a = \frac{1}{8}$

$1 - \left(\frac{1}{8}\right)*6 = \frac{2}{8} = b$

$f\left(x\right) = \frac{1}{8}x^2 + \frac{2}{8}x + 1$

$f\left(8\right) = \frac{1}{8}*8^2+ \frac{2}{8}*8+ 1$

$= 8+2+1 = 11$

f(0)=1, f(4)−f(2)=2, f(6)−f(4)=3

f(x)=Ax2 +Bx +C

f(0)=1

Ax2+Bx+C=1

0+0+C=1

C=1

f(4)−f(2)=2

16A+4B-(4A+2B)=2

12A+ 2B =2 *Eq. 1

f(6)−f(4)=3

36A+6B-(16A+4B)=3

20A+ 2B =3 *Eq. 2

Eq1&2

20A+2B=3

12A+2B=2

A=1/8

B=2/8 or 1/4

f(x)=1/8x2+1/4x+1

f(8)=8+2+1

f(8)=11

Let $f(x) = ax^{2} + bx + c$ . These equations implies

$1. c = 1$

$2. 12a + 2b = 2$

$3. 20a + 2b = 3$

Solving these equations gives $a = \frac{1}{8}, b = \frac{1}{4}, c = 1$ .

So, $f(8) = 11$ .

Our equation will be in the form $f(x) = ax^2 + bx + c$ , since $f(0) = 1$ , $c = 1$ Now with the last two of the three statements about f, we can set up a system of equations $12a + 2b = 2$ $20a + 2b = 3$ solving this we get $a = \dfrac{1}{8}, b = \dfrac{1}{4}$ , now we have $f(x) = \dfrac{1}{8}x^2 + \dfrac{1}{4}x + 1$ $f(8) = 8 + 2 + 1 = 11$

Since it's a quadratic function, we put it in the form f(x) = $ax^{2}$ + bx + c.

And f(0) = 1, so c=1.

Next, f(4) = 16a + 4b + 1

f(2) = 4a + 2b + 1

And their difference would be 12a + 2b which equals 2

Next, f(6) = 36a + 6b + 1

and f(4) = 16a + 4b + 1

and their difference is 20a + 2b which equals 3.

So since we now have that 12a + 2b = 2 and 20a + 2b = 3 , we then solve these two equations to get that a = 1/8 and b = 1/4

So our final equation is thus f(x) = $1/8x^{2}$ + 1/4x + 1 , so to get f(8), what's left to do is to substitute 8 for x in the equation. So we get 8+2+1 = 11.

Let f(x)=ax^2+bx+c if f(0)=1, then 1=a(0)^2+b(0)+c; 1=c if f(4)-f(2)=2, then 2=a(4)^2+b(4)+c-(a(2)^2+b(2)+c); 1=6a+b if f(6)-f(4)=3, then 3=a(6)^2+b(6)+c-(a(4)^2+b(4)+c); 3=20a+2b

We now have c=1; 6a+b=1 and 20a+2b=20. Solving for a and b, we have a=1/8 and b=1/4. Therefore, f(x)=(1/8)x^2+(1/4)x+1 and f(8)=(1/8)(8)^2+(1/4)(8)+1=8+2+1=11

let f(x)=ax^2+bx+c Since f(0)=1 ,c=1 and f(x)=ax^2+bx+1 f(4)-f(2)=12a+2b+1=2 equation-1 f(6)-f(4)=20a+2b+1=3 equation-2 solving 1&2 gives a=1/8 &b=1/4 f(8)=64a+8b+1=8+2+1=11

f(0)= 1

$ax^{2} + bx + c = 0$ so c = 1

f(4) - f(2) = 2

$16a + 4b + 1 - 4a - 2b - 1 = 2$

12a + 2b = 1 .......... (1)

f(6) - f(4) = 3

$36a + 6b + 1 - 16a - 4b - 1 = 3$

20a + 2b = 3 ............(2)

on subtracting .......(1) from .......(2)

8a = 1

a = (\frac{1}{8})

so b = 1- $\frac{6}{8}$ = $\frac{1}{4}$

f(8) = $\frac{1}{8} *8^{2} + \frac{1}{4}*8 + 1$ = 8 + 2 + 1 = 11