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Making the equation of a parabola

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    So i have a parabola graphed, i need to express y in terms of X. The vertex is 2,10 and the roots are 0,0 and 4,0.

    2. Relevant equations


    3. The attempt at a solution

    I thought of working the quadratic in a reverse process, but it doesn't work. Because one of the roots is at 0,0, i guess it doesnt have either b or c. Because the paraboal is upwards and not downards, i guess X must be negative.
  2. jcsd
  3. Aug 10, 2011 #2


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    Use the vertex form of the quadratic:
    y = a(x - h)^2 + k,
    where (h, k) is the vertex and a is the same as the a in y = ax^2 + bx + c. Then plug in either root for x and y to solve for a. Once you get the complete vertex form, change it standard form and you'll see what b and c are.
  4. Aug 10, 2011 #3
    man im feeling dumb, how am i supposed to know what b and c are?

    i got to the point where i reachd 2b+c=9.52 after doing some algebra, with the other root, i have another result, so im also confused by that.
  5. Aug 10, 2011 #4


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    There are three possible equations for a parabola

    y = ax2 + bx + c
    y = a(x-e)2 + f [we don't usually use e and f for constants]
    y = a(x-u)(x-v)

    Any of them can be used, along with the co-ordinates of any 3 points on the graph, to generate 3 simultaneous equations in 3 unknowns , which can then be solved.

    The second is quite useful if you happen to know the Turning Point - you can evaluate some of the constants by inspection.

    The third is particularly useful if you know the roots [x intercepts] as again you can solve some constants by inspection.
  6. Aug 10, 2011 #5

    Ray Vickson

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    If f(x) = a*x^2 + b*x + c is a quadratic with roots r1 and r2, it must be of the form k*(x - r1)*(x - r2), because it must vanish at x = r1 and at x = r2, and have powers of x no higher than 2.

  7. Aug 11, 2011 #6


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    Not sure how you got that. Let me try again. Start with the vertex form:
    y = a(x - h)^2 + k

    The vertex is (2, 10), so you plug that in for h and k:
    y = a(x - 2)^2 + 10

    Notice that there are still 3 variables. But we are given two points: (0, 0) and (4, 0). So plug in either x = 0 & y = 0, or x = 4 & y = 0, into the equation above. Now the only unknown is a. Solve for a.

    Take this equation below and now plug in the value for a:
    y = a(x - 2)^2 + 10

    Rewrite the above into standard form (y = ax^2 + bx + c) by squaring the binomial in the parentheses and simplifying. At the end, the number in front of x will be your b value, and the constant will be your c value. Now try it.
  8. Aug 11, 2011 #7


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    vertex is 2,10 and the roots are 0,0 and 4,0.
    Putting x= 2, y= 10, x= 0, y= 0, and x= 4, y= 4 into [itex]y= ax^2+ bx+ c[/itex] gives you three linear equations for a, b, and c.
  9. Aug 11, 2011 #8


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    Perhaps you could show us how you managed that.
  10. Aug 15, 2011 #9
    well, i guess this was the most useful of the posts, thanks
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