008A Sample Final A, Question 3

Question: a) Find the vertex, standard graphing form, and X-intercept for $x=-3y^{2}-6y+2$

                  b) Sketch the graph. Provide the focus and directrix.

Note: In this problem, what is referred to as standard graphing form is the vertex form, in case you search on the internet.

Foundations
1) What type of graph is this? (line, parabola, circle, etc.)
2) What is the process for transforming the function into the standard graphing form?
3) After we have the standard graphing form how do you find the X-intercept, and vertex?
4) Moving on to part b) How do we find a $3^{rd}$ point on the graph?
5) From the standard graphing form how do we obtain relevant information about the focus and directrix?
Answers:
1) The function is a parabola. Some of the hints: We are asked to find the vertex, and directrix. Also only one variable, of x and y, is squared.
2) First we complete the square. Then we divide by the coefficient of x.
3) To find the X-intercept, replace y with 0 and solve for x. Since the parabola is in standard graphing form, the vertex of $x-h=a(y-k)^{2}$ is (h, k).
4) To find a $3^{rd}$ point, we can either use the symmetry of a parabola or plug in another value for x.
5) From the equation $x-h=a(y-k)^{2}$, we use the equation $a={\frac {1}{4p}}$ to find p. P is both the distance from the vertex to the focus and the distance from the vertex to the directrix.

Solution:

Step 1:
There are two ways to determine the standard graphing form.
Regardless of the method the first step is the same: subtract 2 from both sides to yield $x-2=-3y^{2}-6y$
Method 1:
Divide both sides by -3 to make the coefficient of $y^{2}$, 1. This means ${\frac {-1}{3}}(x-2)=y^{2}+2y$
Complete the square to get ${\frac {-1}{3}}(x-2)+1=(y^{2}+2y+1)=(y+1)^{2}$
Multiply both sides by -3 so $(x-2)-3=-3(y+1)^{2}$, and simplify the left side to yield $x-5=-3(y+1)^{2}$
Method 2:
Instead of dividing by -3 we factor it out of the right hand side to get $x-2=-3(y^{2}+2y)$.
Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in $x-5=-3(y^{2}+2y+1)=-3(y+1)^{2}$
We mention here that some instructors/professors are particular about which of these two methods you use.

Step 2:
Since the parabola is in standard graphing form we can read off the vertex, which is (5, -1).
We get the X-intercept by replacing y with 0 and solving for x. So $x-5=-3(1)^{2}$, and the X-intercept is (2, 0).

Step 3:
Now we need the value of p using the relation $a={\frac {1}{4p}}$, where a = -3.
So $-3p={\frac {1}{4}}$, and $p={\frac {1}{-12}}$.

Step 4:
Since $a<0$, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at $(5-{\frac {1}{12}},-1)=({\frac {59}{12}},-1)$.
We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is $x=5+{\frac {1}{12}}={\frac {61}{12}}$

Final Answer:
Vertex: (5, -1), standard graphing form: $x-5=-3(y+1)^{2}$, X-intercept: (2, 0), focus: $({\frac {59}{12}},-1)$, directrix: $x={\frac {61}{12}}$

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