1. ## constructing a graph

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

2. 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?

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

4. 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

5. 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

6. thanks a lot guys
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

7. Originally Posted by Raoh
thanks a lot guys
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)

8. Originally Posted by Raoh
thanks a lot guys
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

9. 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.

10. 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

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

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

01

13. oooops ! sorry.. my mistake ,in your previous post you wrote :
$Ax + By = C$,the $C$constant depends in $A$and $B$,right ?

14. ok,thanks guys i got it