Finding the vertex, y-intercept and axis of symmetry

[MammaOrnelas]

I am lost and confused. I have been on the same problem for 2 hours. I know all the formulas, but I'm not doing something right...

 

[Unknown008]

For a general equation $$y = a(x-b)^2 + c$$, the vertex is (b, c), the axis of symmetry x = b and the y-intercept is obtained by setting x = 0.

That form of a quadratic equation is called the completed square form.

 

[MammaOrnelas]

Is it okay if you call me? I have soooooooooooooooo much homework left to do and my final is tomorrow. I have been stuck for the last 5 hours...

 

[Unknown008]

I'd love to... but I'm from the other side of the world and I doubt I can afford the telephone bills (Worried)

I'd be very happy to help you as much as I can here, should you have any problems.

 

[MammaOrnelas]

I think it will be easier on both parts if you can talk me through it.

---------- Post added at 23:09 ---------- Previous post was at 23:07 ----------

the problem is f(x)= -3(x-6)^2-4 I know I have to set it equal to zero. It wants the vertex, y intercept and axis of symmetry. I don't know how to do this.

I had to miss three days of class last week due to a death of a family member in another state. I am extremely lost.....

 

[Unknown008]

Okay, from what I gave you earlier, can you see that:

a = -3
b = 6
c = -4

?

 

[MammaOrnelas]

sorta...i know that when its ex: x^2+4x+8, i know which ones are a, b, and c. But I don't see how you could tell which one was a b and c in that problem. Do I have to solve the problem first?

 

[Unknown008]

Not at all! And you'll see that it's just like your example x^2 + 4x + 8.

That one general form is: $$y = ax^2 + bx +c$$
Your example: $$y = x^2 + 4x + 8$$

Which makes: a = 1, b = 4 and c = 8.

In the completed square form, we have:
$$y = a(x-b)^2 + c$$
And in your problem: [tex]y= -3(x-6)^2-4[/tex]

In the same way, we have: a = -3, b = 6 and c = 4.

Does that make things clearer?

And subsequently, remember what I said in my first post here:
Vertex = (b, c)
Axis of symmetry = x = b
y-intercept is obtained by putting x = 0.

 

[MammaOrnelas]

makes a little bit more sense...i was thinking since the (...) part was squared, I had to solve it first... I rememberd that b is what ever makes the (...) = 0 and that was it. Okay...lemme go try and work this problem. It's on mymathlab.com and I hate it! Thanks for the tips...brb

---------- Post added at 23:30 ---------- Previous post was at 23:26 ----------

I got all of it right but it said that the y-intercept is (0,-112)...how did they get that

 

[Unknown008]

Okay, let's see:

[tex]y = -3(x-6)^2-4[/tex]

By putting x = 0, we get:

[tex]y = -3(0-6)^2-4[/tex]

[tex]y = -3(36)-4[/tex]

[tex]y = -108-4[/tex]

[tex]y = -117[/tex]

Wasn't that easy? (Wink)

 

[MammaOrnelas]

I see now...i was making the whole parethense 0 and getting -7 for my answer...Okay...it gave me a new problem...let me try it!