Finding Parabola Equation with Vertex and X-Intercept

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Homework Help Overview

The discussion revolves around finding the equation of a parabola given its vertex and one x-intercept. The vertex is specified as (0, -1920) and the x-intercept as (96, 0). There is also a later correction regarding the vertex and x-intercept values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the general form of a parabola and how it relates to the vertex form. There are questions about which constants in the standard quadratic equation may be zero based on the given vertex and intercepts. Some participants express confusion about the terminology used in the problem statement.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided guidance on how to express the equation in vertex form, while others are questioning the clarity of the problem statement itself. There is no explicit consensus on the correct approach yet.

Contextual Notes

There are noted discrepancies in the values provided for the vertex and x-intercept, leading to confusion. Participants are also grappling with the implications of the problem statement regarding the use of "one of the points" versus the x-intercept.

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Homework Statement



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=)
 
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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?
 
umm... :redface: .. not sure wht ur talkin about.. :confused:
olgranpappy said:
the three constants happens to be zero. Which one?
 
Last edited:
What is the general form of a quadratic equation?

How is it written when we want to emphasize what the vertex is?
 
Last edited:
well the general form of a parabola that is emphasized in vertex form is
y=a(x+h)^2+k
 
yo0o0ogii said:
well the general form of a parabola that is emphasized in vertex form is
y=a(x+h)^2+k
It's actually ...

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

Ok so, we know that h is the apex [tex]h=-\frac{b}{2a}[/tex] 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 ...

[tex]y=a(x-h)^2+k \rightarrow y=ax^2-1920[/tex]

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, [tex]P(\pm 96,0)[/tex], what we want to know now is the value for a ...

Let's see your answer.
 
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k i made a huge mistake

1. Homework Statement

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
 
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.
 
yo0o0ogii said:
k i made a huge mistake

1. Homework Statement

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
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
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
HallsofIvy said:
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.

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