My attempt:
$$f(x) = a(x - r)(x - s)$$ $$f(x) = a(x-(2 + \sqrt {3}))(x-(2- \sqrt {3}))$$
From here, I'm stuck. I can't remember where to go from this point and need some help.
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What is $f(2)$ and what should it be? – Hagen von Eitzen Jan 26 '16 at 16:40
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Hint: $$\text{Since }(2,5)\text{ belongs to the parabola it follows } f(2)=5$$ – Ángel Mario Gallegos Jan 26 '16 at 16:41
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$f(2)=5$, so $a=\frac{5}{(2-(2+\sqrt{3}))(2-(2-\sqrt{3}))}$ – user236182 Jan 26 '16 at 16:41
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You have $2+\sqrt 2$ in the title, but $2+\sqrt 3$ in the body. – TonyK Jan 26 '16 at 16:49
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3Also, note that you can simplify $(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))$ into a rational polynomial and it's probably worth expanding that out before you go further. – Steven Stadnicki Jan 26 '16 at 19:31
2 Answers
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You've used several pieces of your information, but not all. You haven't used the fact that it passes through $(2,5)$, i.e. that $f(2) = 5$. So you should plug in $2$ and $5$ and then solve for $a$.
Future
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The vertex must lie on the axis of symmetry. This is equal to the mean of the $x$-coordinates of the roots, or $x = \frac{1}{2} \left(2+\sqrt3 + 2-\sqrt3 \right) = 2$.
Therefore we can write the equation of the parabola in vertex form: $y = a(x-2)^2 + b$. Use the fact that it passes through $(2,5)$, and $(0, 2+\sqrt{3})$ to form a system in two equations:
$$ \left\{ \begin{array}{c} a(2 - 2)^2 + b = 5 \\ a(2 + \sqrt3 - 2)^2 + b=0\\ \end{array} \right. $$
Now the equation of the parabola is obvious.
Toby Mak
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