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## rosedewittbukater Group Title Help and explanation? What are the vertex, focus, and directrix of the parabola with the given equation? 12y = x^2 –6x + 45 (1 point) one year ago one year ago

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1. rosedewittbukater Group Title

vertex (–3, –3); focus (0, 3); directrix y = 3 vertex (3, 3); focus (3, 6); directrix y = 0 vertex (3, –3); focus (0, 6); directrix y = –3 vertex (–3, 3); focus (3, –3); directrix y = 0

2. rosedewittbukater Group Title

What is an equation of a parabola with the given vertex and focus? vertex: (5, 4) ; focus: (8, 4) (1 point) x =(y – 4)2 + 5 y =(x + 4)2 – 5 x =(y + 4)2 – 5 y =(x – 4)2 + 5

3. rosedewittbukater Group Title

@perurabo do you know how to do this? For the first one I tried to get it into standard form and complete the square for vertex form, but got confused with the fractions :/

4. rosedewittbukater Group Title

@whpalmer4 @wio ???

5. whpalmer4 Group Title

The vertex is the bottom of the "bowl" (if the parabola opens up) or the top of the "dome" (if the parabola opens down). There are a number of approaches you could use, especially given that you have a multiple-choice question here. Obviously, you can eliminate any answer for which the given point doesn't satisfy the equation of the parabola. Given that all of the answer choices are 3 or -3 for the value of $$x$$, and the equation both squares $$x$$ (giving a positive number) and adds 45, there's no question that the value of the right hand side of the equation is going to be positive, which means $$y$$ is also positive. That leaves two choices, which you can easily check to see which point is on the curve and which is not. You could also manipulate the equation to give the so-called "vertex form": $y=a(x-h)^2+k$where $$(h,k)$$ is the vertex. Be careful of the signs! Rather than trust my memory, I always figure this out by translation: the simplest parabola $$y=f(x) = x^2$$ has a vertex of $$(0,0)$$. If I want to shift the parabola along the x axis, I add or subtract from the argument to the function: subtracting 1 from the argument means the vertex happens at $$x=1$$ instead of $$x = 0$$, so the shifted parabola is $$y=f(x-1) = (x-1)^2$$. If I wanted to shift the parabola up or down, I add or subtract from the result of the function, so adding 2 to the function will give me a parabola that is shifted up 2. My "test" parabola would thus be $$y = (x-1)^2+2$$ and if we look at the vertex form, we see that gives a parabola with vertex $$(1,2)$$, just as expected. Your parabola is $$12y = x^2-6x+45$$ so some algebra is required to get it into vertex form. I would observe that $(x-3)^2 = (x-3)(x-3) = x^2-3x-3x+9=x^2-6x+9$and from that you should be able to regroup the formula to give you vertex form. Pull out the $$(x-3)^2$$ first, then divide both sides by 12. Yet another way to get the vertex: put your parabola in the form $$y = ax^2 + bx + c$$. The x value of the vertex will be at $$x = -\dfrac{b}{2a}$$ and you can find the y value by plugging the x value into the formula. Again, be careful of the signs! Focus and directrix: for the first problem, you really only have to find the vertex and you've found the focus and directrix, thanks to the format of the question. However, it is useful to know how to do it when you actually need to! If you rearrange the formula so that it has the form $(x-h)^2 = 4p(y-k)$ then the focus will be at $$(h,k+p)$$ and the directrix will be $$y = k-p$$

6. HeLp_Me2 Group Title

What was the right answer?

7. whpalmer4 Group Title

For the first one: $12y = x^2 –6x + 45$Solve for $$y$$ $y = \frac{x^2}{12} -\frac{x}{2} + \frac{45}{12}$ $y = ax^2+bx+c$$a=\frac{1}{12}$$b=-\frac{1}{2}$$c=\frac{45}{12}$ vertex x-coordinate is $x=-\frac{b}{2a} = -\frac{-1/2}{2/12} = 3$ $y = \frac{3^2}{12} -\frac{3}{2} + \frac{45}{12} = 3$ Vertex is at $$(3,3)$$

8. whpalmer4 Group Title

focus and directrix are symmetrically located about the vertex, so focus is at (3,6) and directrix at y = 0

9. HeLp_Me2 Group Title

What about the second one? @whpalmer4

10. HeLp_Me2 Group Title

@whpalmer4 What is an equation of a parabola with the given vertex and focus? vertex: (5, 4) ; focus: (8, 4)

11. HeLp_Me2 Group Title

x= 1/12 (y-4)^2 +5 y= 1/12 (x+4)^2 -5 x= 1/12 (y+4)^2 -5 y= 1/12 (x-4)^2 +5

12. HeLp_Me2 Group Title

@whpalmer4

13. whpalmer4 Group Title

Graph all 4, see if any have the specified vertex and focus.

14. whpalmer4 Group Title

Actually, I'm going to take that back earlier comment. $x = \frac{1}{12}(y-4)^2+5$has the desired properties.

15. whpalmer4 Group Title

Actually, I'm going to take that back earlier comment. $x = \frac{1}{12}(y-4)^2+5$has the desired properties.