Finding Parabola Equation with Vertex and X-Intercept

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To find the equation of a parabola given the vertex (0, 1920) and an x-intercept (960, 0), the equation can be expressed in vertex form as y = a(x - 0)^2 + 1920, simplifying to y = ax^2 + 1920. The value of 'a' can be determined by substituting the x-intercept into the equation, leading to 0 = a(960^2) + 1920. This results in the calculation of 'a' as -1/480. The discussion highlights the confusion around the terminology used for points and intercepts while clarifying the symmetry of the parabola.
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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?
 
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What is the general form of a quadratic equation?

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

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.
 
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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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