constructing a graph

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June 21st 2009, 12:40 PM
Raoh

constructing a graph

hi(Hi)
here's my question :
Construct an example of a polynomial so that its graph goes through the points
$\left ( -1,5 \right )$ and $\left ( 3,-2 \right )$
thanks(Happy)
June 21st 2009, 12:46 PM
skeeter

Quote:

Originally Posted by Raoh

here's my question :
Construct an example of a polynomial so that its graph goes through the points
$\left ( -1,5 \right )$ and $\left ( 3,-2 \right )$

what degree of polynomial?
June 21st 2009, 01:28 PM
Raoh

thanks for your reply,but i don't know what degree is the polynomial,that was the question (Worried).
thanks again.
June 21st 2009, 01:52 PM
skeeter

Quote:

Originally Posted by Raoh

thanks for your reply,but i don't know what degree is the polynomial,that was the question.

o.k. , let's keep it simple. here is a polynomial of degree 1 that works ...

7x + 4y = 13
June 21st 2009, 02:01 PM
yeongil

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

Plug in the vertex for (h, k):
$y = a(x - 3)^2 - 2$

Plug in the other point for (x, y) to solve for a:
$\begin{aligned} 5 &= a(-1 - 3)^2 - 2 \\ 5 &= (-4)^2 a - 2 \\ 5 &= 16a - 2 \\ 7 &= 16a \\ a &= \frac{7}{16} \end{aligned}$

So the equation of the parabola would be
$y = \frac{7}{16}(x - 3)^2 - 2$

01
June 21st 2009, 02:19 PM
Raoh

thanks a lot guys(Happy)
if i wanted to give an example of polynomial of degree 1,it will have always the form :
$7x+4y = k$ and $k$ could be anything,am i right ?
thanks again
June 21st 2009, 02:28 PM
skeeter

Quote:

Originally Posted by Raoh

thanks a lot guys(Happy)
if i wanted to give an example of polynomial of degree 1,it will have always the form :
$7x+4y = k$ and $k$ could be anything,am i right ?
thanks again

no ... not if they pass through the two given points (-1,5) , (3,-2)
June 21st 2009, 02:31 PM
yeongil

Quote:

Originally Posted by Raoh

thanks a lot guys(Happy)
if i wanted to give an example of polynomial of degree 1,it will have always the form :
$7x+4y = k$ and $k$ could be anything,am i right ?
thanks again

No, k can't be just anything.

To find a linear equation, you need to find the slope:
$\begin{aligned} m &= \frac{y_2 - y^1}{y_2 - y^1} \\ &= \frac{-2 - 5}{3 - (-1)} \\ &= -\frac{7}{4} \end{aligned}$

Use point-slope form:
$\begin{aligned} y - y_1 &= m(x - x_1) \\ y - 5 &= -\frac{7}{4}(x + 1) \\ 4y - 20 &= -7(x + 1) \\ 4y - 20 &= -7x - 7 \\ 7x + 4y - 20 &= -7 \\ 7x + 4y &= 13 \\ \end{aligned}$

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

01
June 21st 2009, 02:33 PM
Raoh

Quote:

Originally Posted by skeeter

no ... not if they pass through the two given points (-1,5) , (3,-2)

of course they have to pass through the two points $(1.5),(3.-2)$ sorry i forgot to mention that.does this make my previous post right ?
thanks.
June 21st 2009, 02:40 PM
yeongil

Quote:

Originally Posted by Raoh

of course they have to pass through the two points $(1.5),(3.-2)$ sorry i forgot to mention that.does this make my previous post right ?
thanks.

The answer is still no. A linear equation in standard form is
$Ax + By = C$
where A, B, and C are constants. Depending on the two given points, you'll have different sets of values for A, B, and C.

01
June 21st 2009, 02:45 PM
Raoh

i don't get it (sorry(Itwasntme)),if a line passes by those two points,it will have always the same slope $(m = \frac{-7}{4} )$ ,right ?
thanks for helping me.(Happy)
June 21st 2009, 02:47 PM
yeongil

Quote:

Originally Posted by Raoh

i don't get it (sorry(Itwasntme)),if a line passes by those two points,it will have always the same slope $(m = \frac{-7}{4} )$ ,right ?
thanks for helping me.(Happy)

Yes.

01
June 21st 2009, 02:54 PM
Raoh

oooops ! sorry.. my mistake (Itwasntme),in your previous post you wrote :
$Ax + By = C$ ,the $C$ constant depends in $A$ and $B$ ,right ?
June 21st 2009, 03:26 PM
Raoh

ok,thanks guys i got it (Wink)
(Clapping)