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Thread: Parabela

  1. #1
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    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)
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  2. #2
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    (-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: $\displaystyle y = 0 = a{\color{red}(x - r )}{\color{blue}(x-s)}$

    The only way that your equation will yield 0 is if either the red or blue is equal to 0 (remember, $\displaystyle a \times b = 0$: 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.
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    equation to x intercept

    My equation is now y=2x2-5x-12. How do i get x intercepts from this
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    Set y = 0 and factor your expression.

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

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

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

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

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

    Now apply similar reasoning to your question.
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  5. #5
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    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?
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  6. #6
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    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
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  7. #7
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    There are many ways to factor trinomials with a coefficient other than 1 in front of the $\displaystyle x^{2}$ term.

    I like doing it this way. For the general trinomial $\displaystyle ax^2 + bx + c$, we find two numbers such that they multiply to $\displaystyle a \times c$ and add up to $\displaystyle b$. Let's call these two numbers $\displaystyle u$ and $\displaystyle v$. 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: $\displaystyle 3x^{2} + 14x + 15$

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

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

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

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

    Voila. See if you can apply this to your expression.
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