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Math Help - Parabola Equation

  1. #1
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    Exclamation Parabola Equation

    Is there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
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  2. #2
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    Quote Originally Posted by nath_quam
    Is there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
    Hello,

    if the vertex is V(x_v \vert y_v) then you can use the vertex-form of the parabola:
    p(x)=a\cdot (x-x_v)^2+y_v

    Plug in the coordinates of the vertex:

    p(x)=a\cdot (x-0)^2+1. Now is left only the coefficient a. Plug in the coordinates of the point (I've named it A):

    3=a\cdot (1-0)^2+1 and you'll get a = 2.

    So the parabola is: p(x)=2\cdot (x)^2+1

    I've attached a diagram.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Parabola Equation-par_durch_pkt.gif  
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  3. #3
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    nath_quamIs there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)

    Hi Nath_Quam: In short, yes, there are infinitely many such parabolas that meet the conditions given in your narrative. To begin, we write your parabola in functional notation, i.e., f(x) = ax^2 + bx + c. [Note: There is nothing compelling us to use f(x). We could just as legitimately have written g(x)=..., h(x)=..., or, simply y= ax^2 + bx + c. All are quadratic equations and, hence, have parabolic graphical representations].

    Now, recall that any parabola has x-value: -b/2a. According to your given data, the vertex is at (1,3). Hence 1 =-b/2a, and we immediately observe b=0 in function, f(x) = ax^2 + bx + c. So, f(x) = ax^2+c.
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  4. #4
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    Quote Originally Posted by Rich B.
    nath_quamIs there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
    ...
    Now, recall that any parabola has x-value: -b/2a. According to your given data, the vertex is at (1,3). Hence 1 =-b/2a, and we immediately observe b=0 in function, f(x) = ax^2 + bx + c. So, f(x) = ax^2+c.
    Hello,

    I understood that the vertex is V(0,1). Did I miss something?

    Greetings

    EB
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  5. #5
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    Hello, nath_quam!

    Is there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
    Assuming it is a "vertical" parabola, it has the form: . y\:=\:ax^2 + bx + c


    The vertex of a parabola is at: x = \frac{-b}{2a}

    Since this parabola has its vertex at x = 0, we see that b = 0.

    . . The equation (so far) is: . y\:=\:ax^2 + c


    Since (0,1) is on the parabola: . 1\:=\:a\cdot0^2 + c\quad\Rightarrow\quad c = 1

    . . The equation is: . y\:=\:ax^2 + 1


    Since (1,3) is on the parabola: . 3\:=\:a\cdot1^2 + 1\quad\Rightarrow\quad a = 2

    . . Therefore, the equation is: . \boxed{y \:=\:2x^2 + 1}
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  6. #6
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    response to earboth

    Quote Originally Posted by earboth
    Hello,

    I understood that the vertex is V(0,1). Did I miss something?

    Greetings

    EB
    Greetings Earboth:

    Late last night, I began the process of addressing the given inquiry. After completing a few lines, however, circumstances arose such that I was unable to finish that which I had begun. I thought I had clicked the "cancel" icon but...

    This morning, I awoke to read two well articulated, mathematically complete responses to the originator's query. Any contributory words I may have to offer now, would be an inefficient streatch of my stiff, early morning, pre-caffine enhanced cerebrum.

    And that's that. Some few hours from now, I shall delete the initial post, as well as this clarification of the same.

    Enjoy your day.

    Regards,

    Rich B.
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