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"Write the equation of a parabola with zeroes af 4+/- root5 passing through the point (1,-12), in standard form"

I tried but I when i did y= (x - 6.236)(x-1.763) it didn't add up to -12 when i subbed in 1
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That's why you'll need an **extra factor**

    y = A · [x - (4+√5)] · [x - (4-√5)]

You can sub in the point and solve for A
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The polynomial you wrote is a *monic* polynomial, meaning the leading coefficient is 1. The polynomial you're looking for may not be monic.

So you know the roots are 4 + √5 and 4 - √5, so the polynomial looks like p(x) = a(x - 4 - √5)(x - 4 + √5).

We can expand out the product:

    (x-4-√5)(x-4+√5) = x² + (-4+√5)x + (-4-√5)x + (-4-√5)(-4+√5)
                     = x² - 8x + 11

So our polynomial is p(x) = a(x² - 8x + 11).

Now we use the fact the p(1) = -12, so that -12 = a(1² - 8\*1 + 11) = 4a, so a = -3.

The polynomial is p(x) = -3(x² - 8x + 11) = -3x² + 24x - 33.
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How about you keep the roots in that form instead of making them into decimals?

(x - (4 + sqrt(5)))(x - (4 - sqrt(5)))


If you expand it out / consider Vieta's, you'll get x^2 + bx + c, where:

- b will be the negation of the sum of the roots ( -(r1 + r2) )

- c will be the product of the roots ( r1r2 ).  

4 + sqrt(5) and 4 - sqrt(5) are pretty straightforward to add, and also to multiply, especially if you can recognize the fact that they're conjugates ( (a + b)(a - b) = a^2 - b^(2) ).

Once you get some basic parabola, you should be able to scale it by some factor (the other comments address this) so that substituting x gives you -12.
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