Mathematics: Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created? (5 points) f(x) = one twelfth (x − 3)2 − 1 f(x) = − one twelfth (x − 3)2 − 1 f(x) = − one sixth (x + 3)2 + 1 f(x) = one sixth (x − 3)2 + 1

BStep-by-step explanation:I did the test :)...

Using a directrix of y = 2 and a focus of (3, −4), - id234372256, nosleepbrooklyn2006

Ответ:

avonsluv8809

11.03.2022

Check the picture below, so keeping in mind that, the distance "p" from the vertex to either the focus point or directrix is the same, that means the vertex is half-way in between those two folks.

now, the directrix is above, and the focus point is below, that means is a vertical parabola, meaning the "x" is the squared variable, and it opens downwards, meaning the distance "p" is a negative value, in this case is -3.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------$

$\bf \begin{cases}
h=3\\
k=-1\\
p=-3
\end{cases}\implies (x-3)^2=4(-3)[y-(-1)]
\\\\\\
(x-3)^2=-12(y+1)\implies \cfrac{(x-3)^2}{-12}=y+1
\\\\\\
-\cfrac{1}{12}(x-3)^2-1=y$

Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created? (5 points) f

Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created? (5 points) f

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Ответ:

Brook211

02.09.2020

Mathematics

Step-by-step explanation:

I did the test :)

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Using a directrix of y = 2 and a focus of (3, −4), what quadratic function is created? (5 points) f(x) = one twelfth (x − 3)2 − 1 f(x) = − one twelfth (x − 3)2 − 1 f(x) = − one sixth (x + 3)2 + 1 f(x) = one sixth (x − 3)2 + 1

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