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Plz help mid life problem here!!!! Find the standard form of the equation of the parabola with the given characteristic(s) and the vertex at the origin.....??????
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y = x^2
How can u plZ explain.?????
If you plot y against x for y = x^2 When x = 0, y = 0 When x = 1, y = 1 When x = 2, y = 4 When x = -1, y = 1 When x = -2, y = 4 Plot these points on a graph, and you will obtain the required parabola.
Wait the directrix is y=3
Sorry
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The vertex of a parabola of the form \(y=ax^2+bx+c\) is located at the following point.\[\huge \text{Vertex}=\left(-\frac{b}{2a},\ c-\frac{b^2}{4a}\right)\]So, come up with \(a\), \(b\), and \(c\) that gives a vertex \((0,0)\). (really easy)
I honestly do not understand
Sorr:(
Y
Peas help me!
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Please
:(
Plz,!!!!!!!!
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