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What is the vertex for y=3/4x^2?
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Are you allowed drawing the graph of the equation?
yes
Have you done so yet? I can't remember how to do this....
y=3/4x^2 = (3/4)(x-0)+0 or the vertex is at (0,0)
Develop an intuition for the problem: you can tell the vertex is at (0, 0) in less than a second.
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|dw:1326171511911:dw|
Yes, (3/4)x^2 is a multiple of y=x^2, so the vertex does not change.
so vertex would be y= ax^ vertex =a ?
y= ax^2
The "a" will magnify the height of the parabola, but the vertex will remain at (0,0). Think of the vertex equation: y=a(x-h)+k If h and k are zero, then other values of a will not change the vertex, which is at (h,k).
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ahh ok
so is just like the vertex of a triangle kind of :P
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