hi(Hi)

here's my question :

Construct an example of a polynomial so that its graph goes through the points

$\displaystyle \left ( -1,5 \right ) $ and $\displaystyle \left ( 3,-2 \right )$

thanks(Happy)

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- Jun 21st 2009, 12:40 PMRaohconstructing a graph
hi(Hi)

here's my question :

Construct an example of a polynomial so that its graph goes through the points

$\displaystyle \left ( -1,5 \right ) $ and $\displaystyle \left ( 3,-2 \right )$

thanks(Happy) - Jun 21st 2009, 12:46 PMskeeter
- Jun 21st 2009, 01:28 PMRaoh
thanks for your reply,but i don't know what degree is the polynomial,that was the question (Worried).

thanks again. - Jun 21st 2009, 01:52 PMskeeter
- Jun 21st 2009, 02:01 PMyeongil
If you want a quadratic, this is how to do it:

Pick one of the points as the vertex -- though it does matter which one you pick, because you get different answers otherwise, the problem is only asking for an example. Let's pick (3, -2).

The vertex form of the parabola is

$\displaystyle y = a(x - h)^2 + k$

Plug in the vertex for (h, k):

$\displaystyle y = a(x - 3)^2 - 2$

Plug in the other point for (x, y) to solve for a:

$\displaystyle \begin{aligned}

5 &= a(-1 - 3)^2 - 2 \\

5 &= (-4)^2 a - 2 \\

5 &= 16a - 2 \\

7 &= 16a \\

a &= \frac{7}{16}

\end{aligned}$

So the equation of the parabola would be

$\displaystyle y = \frac{7}{16}(x - 3)^2 - 2$

01 - Jun 21st 2009, 02:19 PMRaoh
thanks a lot guys(Happy)

if i wanted to give an example of polynomial of degree 1,it will have always the form :

$\displaystyle 7x+4y = k $ and $\displaystyle k $could be anything,am i right ?

thanks again - Jun 21st 2009, 02:28 PMskeeter
- Jun 21st 2009, 02:31 PMyeongil
No, k can't be just anything.

To find a linear equation, you need to find the slope:

$\displaystyle \begin{aligned}

m &= \frac{y_2 - y^1}{y_2 - y^1} \\

&= \frac{-2 - 5}{3 - (-1)} \\

&= -\frac{7}{4}

\end{aligned}$

Use point-slope form:

$\displaystyle \begin{aligned}

y - y_1 &= m(x - x_1) \\

y - 5 &= -\frac{7}{4}(x + 1) \\

4y - 20 &= -7(x + 1) \\

4y - 20 &= -7x - 7 \\

7x + 4y - 20 &= -7 \\

7x + 4y &= 13 \\

\end{aligned}$

So you see, it**does**matter what the constant term is.

01 - Jun 21st 2009, 02:33 PMRaoh
- Jun 21st 2009, 02:40 PMyeongil
- Jun 21st 2009, 02:45 PMRaoh
i don't get it (sorry(Itwasntme)),if a line passes by those two points,it will have always the same slope $\displaystyle (m = \frac{-7}{4} )$,right ?

thanks for helping me.(Happy) - Jun 21st 2009, 02:47 PMyeongil
- Jun 21st 2009, 02:54 PMRaoh
oooops ! sorry.. my mistake (Itwasntme),in your previous post you wrote :

$\displaystyle Ax + By = C$,the $\displaystyle C $constant depends in $\displaystyle A $and $\displaystyle B$,right ? - Jun 21st 2009, 03:26 PMRaoh
ok,thanks guys i got it (Wink)

(Clapping)