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 7 months ago
How do you graph the quadratic equation 3x^2=3?
I need help, my EOC is next Tuesday.
 7 months ago
How do you graph the quadratic equation 3x^2=3? I need help, my EOC is next Tuesday.

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TurtleMuffin
 7 months ago
Best ResponseYou've already chosen the best response.1First you can divide both sides by 3 to get x^2 = 1. Subtract 1 from both sides and you have x^2  1 = 0. The roots of this are 1 and 1. So this is a standard parabola shifted down one united passing through 1 and 1 on the xaxis.

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0first, what shape does the graph make?

kylewendt
 7 months ago
Best ResponseYou've already chosen the best response.0The graph makes a parabola when completed

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0good, now, that is an important concept, so now, you have one slight problem, This is only in one variable(x) so you can't graph it. Are you sure you copied the question correctly?

kylewendt
 7 months ago
Best ResponseYou've already chosen the best response.0Yes. The exact question is: What are the solutions of the equation 3x^2=3? Use a graph of the related function.

Hero
 7 months ago
Best ResponseYou've already chosen the best response.0@FibonacciChick666, what do you mean it can't be graphed?

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0'graphed' I mean, it's a number line as stated

Hero
 7 months ago
Best ResponseYou've already chosen the best response.0Continuing what Turtlemuffin was saying, you have x^2  1 = 0 which factors to (x + 1)(x  1) = 0

kylewendt
 7 months ago
Best ResponseYou've already chosen the best response.0My question is, after you have your two xintercepts, how do you find the vertex from (x+1)(x1)=0?

Hero
 7 months ago
Best ResponseYou've already chosen the best response.0We can use the vertex formula to graph the vertex. \(x = \dfrac{b}{2a}\)

Hero
 7 months ago
Best ResponseYou've already chosen the best response.0In this case b = 0 and a = 1, so \[x = \frac{0}{2(1)} = 0\]

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0@Hero , one variable= 1D , no?

Hero
 7 months ago
Best ResponseYou've already chosen the best response.0Since \(x = 0\), we can insert 0 into the expression to get y = \(0^2  1 = 1\) so the vertex is \((0,1)\)

kylewendt
 7 months ago
Best ResponseYou've already chosen the best response.0So, the graph would look like this?

TurtleMuffin
 7 months ago
Best ResponseYou've already chosen the best response.1That is an algebraic approach. You should know the graph of y=x^2. It's the standard parabola. We simplified your given equation into 0 = x^2  1. The "minus 1" tells us our standard parabola shifts down one unit. Therefore its xcoordinate remains the same, and it's ycoordinate is shifted down 1 unit from 0, i.e. it's now 1. You can remember all the formulas you want, but if you don't see how an algebraic equation is related to its graphical representation, you won't enjoy math too much. That is has how the graph looks like though!

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0ah, now that I reread the prompt, I understand.

FibonacciChick666
 7 months ago
Best ResponseYou've already chosen the best response.0It wanted the roots, by looking at the related graph

kylewendt
 7 months ago
Best ResponseYou've already chosen the best response.0Thank you all for explaining, as I understand this concept now. My algebra teacher is one of the worst (I know that sounds generic), and intentionally ignores my questions to her on assignments because she doesn't like me. This really helps.
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