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Math Help - constructing a graph

  1. #1
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    Smile 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
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  2. #2
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    Quote Originally Posted by Raoh View Post
    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?
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  3. #3
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    Smile

    thanks for your reply,but i don't know what degree is the polynomial,that was the question .
    thanks again.
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  4. #4
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    Quote Originally Posted by Raoh View Post
    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
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  5. #5
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    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}<br />
5 &= a(-1 - 3)^2 - 2 \\<br />
5 &= (-4)^2 a - 2 \\<br />
5 &= 16a - 2 \\<br />
7 &= 16a \\<br />
a &= \frac{7}{16}<br />
\end{aligned}

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


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  6. #6
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    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
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  7. #7
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    Quote Originally Posted by Raoh View Post
    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)
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  8. #8
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    Quote Originally Posted by Raoh View Post
    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}<br />
m &= \frac{y_2 - y^1}{y_2 - y^1} \\<br />
&= \frac{-2 - 5}{3 - (-1)} \\<br />
&= -\frac{7}{4}<br />
\end{aligned}

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

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


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  9. #9
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    Quote Originally Posted by skeeter View Post
    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.
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  10. #10
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    Quote Originally Posted by Raoh View Post
    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.


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  11. #11
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    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.
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  12. #12
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    Quote Originally Posted by Raoh View Post
    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.


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  13. #13
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    oooops ! sorry.. my mistake ,in your previous post you wrote :
    Ax + By = C,the C constant depends in A and B,right ?
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  14. #14
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    ok,thanks guys i got it
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