8A F11 Q3

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.

Foundations
1) What type of function are we asking you to graph (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 obtain 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 ${\frac {-1}{3}}(x-2)+1=(y^{2}+2y+1)=(y+1)^{2}$
Subtract one from both sides and multiply both sides by -3 so $(x-2)=-3(y+1)^{2}+3$