1. ## Determine eq of quad

Q askes: Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are $2 +\sqrt{3}$ and $2 - \sqrt{3}$

To get started, I'm thinking of factored form $f(x) = a(0 - \sqrt{3}) (0 +\sqrt{3})$

2. You have $f(x)=ax^2+bx+c$ and know that (-4, 5) is on graph of f, then: $16a-4b+c=5$

Indication:
Use Viète's formulas.

3. $\frac{5}{33}(x^2 -4x +1)$ is suppost to be the answer.

However, I cannot seem to get to the answer. I'm using:

$f(x) = a(x -2 \sqrt3)(x -2+\sqrt3)$
$f(x)= a(x+1)(x-5)$
$f(x)=a(x^2 -4x -5)$

... this does not seem to work.

4. Well, probably because $f(x) = a[x -(2 -\sqrt3)][x -(2+\sqrt3)]$, not that expression you wrote.

Now use what you know: (-4, 5) is on the graph of f.

5. Question askes word for word:
Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are $2 +\sqrt{3}$ and $2 - \sqrt{3}$

Wheather wrong or right, the formula below is what I came up with to use.
$f(x) = a[x -(2-\sqrt3)] [(x -(2+\sqrt3)]$

How does it look? It lead me to this: $f(5) = a[-4 -(2 -\sqrt3)] [-4 -(2+\sqrt3)]$

6. If (-4, 5) is on the graph of f, then f(-4)=5. After that calculate.

7. Originally Posted by reallylongnickname
Question askes word for word:
Determine the equation of the quadratic function that passes through (-4,5) if it's zeros are $2 +\sqrt{3}$ and $2 - \sqrt{3}$

Wheather wrong or right, the formula below is what I came up with to use.
$f(x) = a[x -(2-\sqrt3)] [(x -(2+\sqrt3)]$

How does it look? It lead me to this: $f(5) = a[-4 -(2 -\sqrt3)] [-4 -(2+\sqrt3)]$
It's simplest to think of that as a "product of sum and difference": $(a+ b)(a- b)= a^2- b^2$
$f(x= a((x-2)+\sqrt{3})((x-2)- \sqrt{3}= a((x- 2)^2- 3)= a(x^2- 4x+ 4-3)= a(x^2- 4x+ 1)$

But you don't calculate f(5): f(-4)= 5

8. Originally Posted by HallsofIvy
It's simplest to think of that as a "product of sum and difference": $(a+ b)(a- b)= a^2- b^2$
$f(x= a((x-2)+\sqrt{3})((x-2)- \sqrt{3}= a((x- 2)^2- 3)= a(x^2- 4x+ 4-3)= a(x^2- 4x+ 1)$

But you don't calculate f(5): f(-4)= 5
Ahhh, i see. thx. I'm not sure what u mean about not calculating f(5).
When I plug f(5) and x= -4 into your final equation, the result is the same as the answer that was given.

$f(5) = a((-x)^2 -4(-x) +1)$
$f(5) = a((-4)^2 -4(-4) +1)$
$f(5) = a33$
$\frac{5}{33} = a$
$=\frac{5}{33} ((x)^2 -4(x) +1)$