1. ## one variable

Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).

2. Originally Posted by puneet
Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).
let $f(x)$ be a curve. a point $(a,b)$ is on the curve if $f(a)=b$.

if you want a quadratic in one variable, say in $x$, then your it is of the form $f(x)=ax^2 + bx + c$

you are given 3 points. then you can solve $a,b,c$ by solving the system of equation

$f(1)=6$
$f(2)=3$
$f(3)=2$

Thanks

4. ## Is this right

Q)Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).

ANS)Cont.
Thus we have
6 = a + b + c --------(1)
3 = 4a + 2b + c --------(2)
2 = 9a + 3b + c --------(3)
Multiply (1) by 2 and subtract from (2)
Multiply (1) by 3 and subtract from (3)
We get
2a-c= -9 à c-2a=9 --------(4)
And
6a-2c = -16 à 3a-c=-8 --------(5)
Adding (4) and (5) We get
a=1
Putting a in (4) We get
c-2(1)=9 à c-2=9 à c=11
putting a and c in (1)
6=1+b+11 à 6=12+b à b=-6

Hence in order we get
a=1,b=-6,c=11 --------(6)

We have
F(x) =ax2 + bx + c --------(7)

Putting values of f and (6) in (7) we get
x2 - 6x + 5 = 0
x2 - 6x + 8 = 0
x2 - 6x + 9 = 0

Is This right I am getting three equations whats wrong.

5. Originally Posted by puneet
Q)Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).

ANS)Cont.
Thus we have
6 = a + b + c --------(1)
3 = 4a + 2b + c --------(2)
2 = 9a + 3b + c --------(3)
Multiply (1) by 2 and subtract from (2)
Multiply (1) by 3 and subtract from (3)
We get
2a-c= -9 à c-2a=9 --------(4)
And
6a-2c = -16 à 3a-c=-8 --------(5)
Adding (4) and (5) We get
a=1
Putting a in (4) We get
c-2(1)=9 à c-2=9 à c=11
putting a and c in (1)
6=1+b+11 à 6=12+b à b=-6

Hence in order we get
a=1,b=-6,c=11 --------(6)

We have
F(x) =ax2 + bx + c --------(7)

Putting values of f and (6) in (7) we get
x2 - 6x + 5 = 0
x2 - 6x + 8 = 0
x2 - 6x + 9 = 0

Is This right I am getting three equations whats wrong.
your equation is $f(x)=x^2 -6x + 11$.. you should not substitute it on the system..

6. Hello, puneet!

With this type of system of equations,
. . there is a much simpler approach.

. . $\begin{array}{cccc}a + b + c &=& 6 & [1] \\
4a + 2b + c &=& 3 & [2] \\
9a + 3b + c &=& 2 & [3] \end{array}$

Eliminate $c.$
. . $\begin{array}{ccccc}\text{Subtract: [2] - [1]} & 3a + b &=& \text{-}3 & [4] \\
\text{Subtract: [3] - [2]} & 5a + b &=& \text{-}1 & [5] \end{array}$

Eliminate $b.$
. . $\text{Subtract: [5] - [4] }\;\;2a\:=\: 2 \quad\Rightarrow\quad \boxed{a \:=\:1}$

$\text{Substitute into [4]: }\;3 + b \:=\:\text{-}3 \quad\Rightarrow\quad\boxed{ b \:=\:\text{-}6}$

$\text{Substitute into [1]: }\;1 - 6 + c \:=\:6 \quad\Rightarrow\quad\boxed{ c \:=\:11}$

$\text{Therefore: }\;f(x) \;=\;x^2-6x + 11$

Thanks