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Math Help - polynomial functions

  1. #1
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    polynomial functions

    There is not much information given in this problem, nor is there any examples so I am lost.

    Here is the problem:

    Decide if it is possible for a parabola to have the indicated number of x-intercepts. If it is possible, find an equation of such a parabola. If not possible explain why.

    a. One intercept

    b. Three intercepts

    The other part of the problem:

    The vertex of a parabola is (5,3) and the parabola passes through the point (8,10). Find a third point that lies on the parabola?

    I know that the vertex is the lowest point. Is there an equation that I would use to find the third point?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by getnaphd View Post
    There is not much information given in this problem, nor is there any examples so I am lost.

    Here is the problem:

    Decide if it is possible for a parabola to have the indicated number of x-intercepts. If it is possible, find an equation of such a parabola. If not possible explain why.

    a. One intercept
    yes, it is possible

    example: y = x^2

    b. Three intercepts
    no, it is not possible.

    note that the x-intercepts of a polynomial are the roots of the polynomial. and in general, an n^{th} degree polynomial can have AT MOST n roots.

    specific to the case of parabolas:

    a parabola is a polynomial of the form y = ax^2 + bx + c

    the x-intercepts are given by the roots of the equation, namely, the solutions to ax^2 + bx + c = 0

    it can be shown that the roots are given by x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}. This is known as the quadratic formula, it can be derived by completing the square.

    clearly there are three cases:

    case 1: if b^2 - 4ac > 0 we have TWO distinct, real roots.

    case 2: if b^2 - 4ac = 0 we have ONE real root (with a multiplicity of 2).

    case 3: if b^2 - 4ac < 0, we have TWO complex roots

    Thus we see, a quadratic can have at most two roots, which are the x-intercepts

    if your professor demands a more concrete proof, you can look up the derivation of the quadratic formula here. it uses completing the square, which you will learn how to do soon, if you haven't already


    The other part of the problem:

    The vertex of a parabola is (5,3) and the parabola passes through the point (8,10). Find a third point that lies on the parabola?

    I know that the vertex is the lowest point. Is there an equation that I would use to find the third point?
    It's 2am and i'm tired, so maybe there's an easier way to do this than what i'm seeing. but i'd set up a system of three equations.

    indeed, there is a formula for the vertex of a parabola (which by the way can be the highest OR lowest point of the parabola). the formula gives the x-coordinate for the vertex.

    the x-coordinate for the vertex of a parabola of the form y = ax^2 + bx + c is given by: x = \frac {-b}{2a}

    we will use this formula to solve this problem.

    Here's the information we have:

    (5,3) ---vertex
    (8,10) ----a point the parabola passes through

    Let the parabola be of the form y = ax^2 + bx + c

    since we have the vertex is (5,3), we know two things:

    5 = \frac {-b}{2a} --------the vertex formula.

    b = -10a ...........................(1)

    also, when x = 5, y = 3. so we can plug this into our general form to get:

    3 = 25a + 5b + c ................(2)

    since we have (8,10), we know, when x = 8, y = 10. this means

    10 = 64a + 8b + c ..................(3)


    So now we have the system of equations:

    b = -10a ...........................(1)
    3 = 25a + 5b + c ................(2)
    10 = 64a + 8b + c ..............(3)

    From equation (1), we see that b = -10a, substitute this for b in equations (2) and (3) to get two new equations.

    10 = -16a + c ........................(4)
    3 = -25a + c ..........................(5)

    \Rightarrow 7 = 9a .....................(4) - (5)

    \Rightarrow \boxed {a = \frac {7}{9}}

    but, 10 = -16a + c

    \Rightarrow 10 = -16 \left( \frac {7}{9} \right) + c

    \Rightarrow \boxed { c = \frac {202}{9} }

    also, recall that b = -10a

    \Rightarrow b = -10 \left( \frac {7}{9} \right)

    \Rightarrow \boxed { b = - \frac {70}{9} }

    Thus, our quadratic is: y = \frac {7}{9} x^2 - \frac {70}{9} x + \frac {202}{9}

    Wow! That's weird quadratic. Hopefully there are more than one solutions and someone else can find a nicer one
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