Results 1 to 7 of 7

Math Help - Parabela

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    6

    Exclamation Parabela

    I have a parabella with points (-3,0) and (5,0). These are two points are on the x-axis. The bottom point is (1,-4). what would the equation for this be in the form of y=a(x-r)(x-s)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    (-3,0) and (5,0) are on your parabola i.e. if you plug in -3 and -5 into your equation, you'll get 0.

    Now looking at your equation: y = 0 = a{\color{red}(x - r )}{\color{blue}(x-s)}

    The only way that your equation will yield 0 is if either the red or blue is equal to 0 (remember, a \times b = 0: either a or b or both are 0). Since we know x = -3 and 5 will work, we can see that r = -3 and s = 5 (since 3 - (-3) = 0 and 5 - 5 = 0).

    Now all that's left to do is solve for a. You have the point (1, -4) - i.e. if x = 1 and you plug it into your equation, you'll get y = -4. So plug this into your expression from above and you'll get your parabola.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    6

    equation to x intercept

    My equation is now y=2x2-5x-12. How do i get x intercepts from this
    Follow Math Help Forum on Facebook and Google+

  4. #4
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Set y = 0 and factor your expression.

    For example, if your question was:
    Find the x-intercepts of y = x^{2} + 7x + 10.

    Set y = 0 and factor your expression:
    0 = x^{2} + 7x + 10
    0 = (x + 5)(x+2)

    Equate each bracketed term to 0 (like I mentioned earlier if a \times b = 0 then either a or b or both are equal to 0).

    x + 5 = 0 \: \: \Rightarrow \: \: x = -5
    x + 2 = 0 \: \: \Rightarrow \: \: x = -2

    So, your x-intercepts are (-5, 0) and (-2, 0).

    Now apply similar reasoning to your question.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2008
    Posts
    6

    parabella

    there is a 2 infront of the x squared and is making it hard for me to factor 2x2-5x-12. How do i do this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2008
    Posts
    4
    Quote Originally Posted by mcgill33 View Post
    there is a 2 infront of the x squared and is making it hard for me to factor 2x2-5x-12. How do i do this?
    2x^2 - 5x - 12
    = (2x+3)(x-4)
    2x = -3 or x = 4
    x = -3/2 or 4
    sorted
    Follow Math Help Forum on Facebook and Google+

  7. #7
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    There are many ways to factor trinomials with a coefficient other than 1 in front of the x^{2} term.

    I like doing it this way. For the general trinomial ax^2 + bx + c, we find two numbers such that they multiply to a \times c and add up to b. Let's call these two numbers u and v. We split up bx into ux + vx and you should be able to factor it.

    An example would probably help clear this up for you.

    Consider: 3x^{2} + 14x + 15

    We have to find two numbers such that they multiply to 3 \times 15 = 45 and add up to 14. 9 and 5 should come up in your head.

    So, we split up 14x into 9x and 5x: 3x^{2} + 14x + 15 = {\color{red}3x^{2} + 9x} + {\color{blue}5x + 15}

    Now factor each coloured expression as much as you can: {\color{red}3x}{\color{magenta}(x + 3)} + {\color{blue}5}{\color{magenta}(x + 3)}

    Notice the (x+3) in pink? Factor it out of the entire expression and you'll get your factored form: ({\color{magenta}x+3})({\color{red}3x} + {\color{blue}5})

    Voila. See if you can apply this to your expression.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum