SOLUTION: How do you graph the parabola of -12(x+3)=(y-3)^2 that includes the directrix and focus on the graph? Focus = (-6,3) Directrix = x=0 Vertex = (-3,3)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: How do you graph the parabola of -12(x+3)=(y-3)^2 that includes the directrix and focus on the graph? Focus = (-6,3) Directrix = x=0 Vertex = (-3,3)       Log On


   

Question 1204077: How do you graph the parabola of -12(x+3)=(y-3)^2 that includes the directrix and focus on the graph?
Focus = (-6,3)
Directrix = x=0
Vertex = (-3,3)

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

+-12%28x%2B3%29=%28y-3%29%5E2
Focus = (+-6 ,+3 )
Directrix = +x=0
Vertex = (+-3 ,+3 )
Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch.
To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form ++-12%28x%2B3%29=%28y-3%29%5E2 ,++x is the independent variable and +y is the dependent variable (so, you choose it). Choose some values for +x and then determine the corresponding +y -values. Then plot the points and sketch the graph.
since you are already given parabola opening sideways and focus is at (+-6 ,+3 ), means the parabola is opening to the +left and graph will be in II and III quadrant
you are also given two points, use them and use formula to find three more
+-12%28x%2B3%29=%28y-3%29%5E2 ...let +x=-4
+-12%28-4%2B3%29=%28y-3%29%5E2
+y=6.46+ or y=-0.46
+-12%28x%2B3%29=%28y-3%29%5E2 ...let +x=-10
+-12%28-10%2B3%29=%28y-3%29%5E2
+y=12.16 or -6.16
x+ |y
-6 |3
-3 |3
-4 |6.46
-4 |-0.46
-10 |12.16
-10 |-6.16

plot these points and sketch the graph