Standard Form Parabolas?(Quadratic equations)

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
jchen24
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Standard Form Parabolas?(Quadratic equations)

Postby jchen24 » Wed Aug 01, 2012 8:07 pm

Hey... I'm signed up because I need help on this parabola questions! :confused:
Here is my problem:
x^+3x-4=6
x^ is like x squared btw:)
Please help me! :D
Last edited by jchen24 on Thu Aug 02, 2012 3:14 pm, edited 1 time in total.

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stapel_eliz
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Postby stapel_eliz » Wed Aug 01, 2012 10:32 pm

jchen24 wrote:Here is my problem:
x^+3x-4=6

What are you supposed to do with this equation? Graph it? Find the domain? Find the range? Find the intercepts? Solve using the Quadratic Formula? Find the directrix, focus, or vertex? Or something else?

When you reply, please show what you've tried so far. Thank you! :wink:

jchen24
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Re:

Postby jchen24 » Thu Aug 02, 2012 3:27 pm

stapel_eliz wrote:
jchen24 wrote:Here is my problem:
x^+3x-4=6

What are you supposed to do with this equation? Graph it? Find the domain? Find the range? Find the intercepts? Solve using the Quadratic Formula? Find the directrix, focus, or vertex? Or something else?

When you reply, please show what you've tried so far. Thank you! :wink:


I'm supposed to graph the equation and find the vertex.
I've tried doing this to find the vertex:
x^+3x-4=6
x^+3x-10=y <----- Not sure if this is right
Then I did the axis equation and got -3/2.
(-3/2)^+3(-3/2)-10=y
When I solved it, I got -12.25. So for the vertex I got (-3/2, -12.25).

Then I tried a different way to find the vertex:
x^+3x-4=6
y=x^+3x-4
Then I did the axis equation again and got 2.
y=(2)^+3(2)-4
y=6
So for the vertex I got (2,6). The thing is that this is actually one of the points on the parabola... ?
I don't know what I did wrong.
Btw: thanks for replying! :D

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maggiemagnet
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Re: Standard Form Parabolas?(Quadratic equations)

Postby maggiemagnet » Thu Aug 02, 2012 7:53 pm

When you graph the two different equations it makes sense that you get two different graphs with two difference vertexes. Try sticking with the first one. They gave you x^2 + 3x - 4 = 6. You can solve this for the zeroes, but this is an equation in one variable so you can't graph it or find the vertex. But you can get this into "equals zero" form like they explain here and solve that for the zeroes. The vertex will be on the axis and is midway between the zeroes. So you can use the "equals zero" form to find the other stuff by replacing the zero with "y". Then you can graph it and you can complete the square or use the formula for the vertex.


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