Question:
Determine the quadratic function that has the given roots (x-intercepts) and passes though the given point.
$x=2+\sqrt{5}, x=2 - \sqrt{5}$. The given point is $(2,10)$.
This is my answer: $y=2(x-2+√5)(x-2-√5)$
Is this correct? In the handout it says that the $a$ value should be 4, which I think is incorrect.
Since they're roots it means that $y=0$ when $x=2\pm \sqrt5$. Therefore, you're looking for an equation of this sort:
$$y=\alpha (x-2-\sqrt5)(x-2+\sqrt5)$$
Because when $x$ is equal to these values, the entire expression will equal $0$.
Now the next thing is the given point. That will determine the value of $\alpha$. Since $y=10$ at $x=2$:
$$10=\alpha(2-2-\sqrt5)(2-2+\sqrt5)=-5\alpha\\\therefore \alpha=-2$$
And so you have:
$$y=-2(x-2-\sqrt5)(x-2+\sqrt5)$$
Which is a quadratic function (if you were to FOIL it out).