# Homework Help: Parabolas Help

1. Jan 8, 2008

### yo0o0ogii

1. The problem statement, all variables and given/known data

How do you figure out the equation of a parabola by only knowing the vertex and ONE of the x intercepts

Vertex: (0, -1920)
X intercept: (96,0)

Help=)

2. Jan 8, 2008

### olgranpappy

In general a parabola is of the form Ax^2 + Bx + C... but in your case one of the three constants happens to be zero. Which one?

3. Jan 8, 2008

### yo0o0ogii

umm... .. not sure wht ur talkin about..

Last edited: Jan 8, 2008
4. Jan 8, 2008

### rocomath

What is the general form of a quadratic equation?

How is it written when we want to emphasize what the vertex is?

Last edited: Jan 8, 2008
5. Jan 8, 2008

### yo0o0ogii

well the general form of a parabola that is emphasized in vertex form is
y=a(x+h)^2+k

6. Jan 8, 2008

### rocomath

It's actually ...

General form of a quadratic equation: $$y=ax^2+bx+c$$ & $$y=a(x-h)^2+k$$

Ok so, we know that h is the apex $$h=-\frac{b}{2a}$$ of our parabola. Since our vertex has points (0,-1920), we know that h=0 b/c of the fact it lies on the y-axis and that b=0 b/c in order for h to equal 0, b=0 ... if a equaled 0, then it would be undefined.

So this reduces our standard quadratic equation to ...

$$y=a(x-h)^2+k \rightarrow y=ax^2-1920$$

Further, since the apex of our parabola lies on the y-axis, we also know another zero, which is -96. Anyways, we can use either one since a parabola is symmetric with respect's to it's apex. Using our givens, $$P(\pm 96,0)$$, what we want to know now is the value for a ...

Last edited: Jan 8, 2008
7. Jan 9, 2008

### yo0o0ogii

k i made a huge mistake

1. The problem statement, all variables and given/known data

How do you figure out the equation of a parabola by only knowing the vertex and ONE OF THE POINTS ( not x intercept)

Vertex: (0, 1920)
X intercept: (960,0)

its actaully POSITIVE 1960 and 960 not 96

8. Jan 9, 2008

### rocomath

Well that point is an x-intercept b/c where exactly does P(960, 0) lie?

Just follow what I told you in Post #6 and it's solved.

9. Jan 10, 2008

### HallsofIvy

Now it makes no sense at all. Why would they say "ONE OF THE POINTS (not x intercept)" and then give you the x intercept?

In any case, you are told that the vertex is at (0, 1920) so you know the equation is of the form y= a(x- 0)2+ 1920= ax2+ 1920. You only need to determine the single number, a. You also know that (960, 0) is a point on the parabola: that is, when x= 960, y= 0. Put those values of x and y into y= ax2+ 1920 and solve the equation for a.

10. May 20, 2009

### darkfire313

Ever heard of the line of symmetry?
A parabola has one in the middle.

1920 - 960 = 960
1920 + 960 = 2780

There! Two X-intercepts!

11. May 20, 2009

### darkfire313

0 = a921600 + 1920. Huh. It means a is -1/480. Wow. Amazing. And how come I am editting this after a year?

Last edited: May 20, 2009