one variable

**puneet** · August 23rd 2008, 02:38 AM

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

**kalagota** · August 23rd 2008, 02:43 AM

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

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

**puneet** · August 23rd 2008, 03:00 AM

Thanks

**puneet** · August 23rd 2008, 03:31 AM

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.

**kalagota** · August 23rd 2008, 03:44 AM

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..

**Soroban** · August 23rd 2008, 04:22 AM

Hello, puneet!

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

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

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

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

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

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

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

**puneet** · August 23rd 2008, 04:32 AM

Thanks

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