Question

Suppose the trajectory of a launched object can be modeled by the parabola with the equation
y = -1/47 (x + 3)^2 +9, where x represents the time after the object was launched and y represents the altitude.

a. Find the vertex, focus, and the directrix of the parabola. Show your work.
b. Graph the parabola.

c. Describe a reasonable domain and range for the graph. Explain your answer.

Answer 1

Strangely fortunate, aren't we? Your parabola equation is already written in what's probably its most convenient form ^_^

Now, you will please compare it with this "standard" form:
\[\Large y = \color{red}p(x-\color{green}a)^2 + \color{blue}b\]
and tell me the values corresponding to \(\color{red}p\), \(\color{green}a\), and \(\color{blue}b\).

Hop to it now... ;)

Answer 2

I mean, for instance, if I had
\[\Large y = \color{red}{12}(x\color{green}{+5})^2+\color{blue}9\]
Then my values would be as follow:
\[\Large \left.\begin{matrix}\color{red}p&= &12\\\color{green}a&=&-5\\\color{blue}b&=&9\end{matrix}\right.\]

Answer 3

p=-1/47, a=-3, b=9

Answer 4

Brilliant.
Now, for your first question, the vertex is simply the point (a,b)
So the vertex is...?

Answer 5

(-3, 9)

Answer 6

Right.
The focus and directrix are a *little* tricky, first you need to find exactly how far they are from the vertex.

And that's determined by your value for p.

Specifically, one quarter of p.

So, what is \(\Large \frac14\) of p?