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Posted

Do a table of plots. So plug 1 into "x" then 2 into "x" and so on and so on.... Then plot these x and y values as coordinates on an xy graph. Then you will have a parabola, then just shift every point down 3 and this 2nd parabola will be "f(x)=2x^2-3"(your equation you were on about).:)

Posted

That isn't a parabola; it's an expression. However, y=2x2-3 is a function, so it can be graphed(and it is a parabola). Make a chart of values with two columns(one side is x, the other y). Choose some x values and then plug them into the equation to get the y values. Some values you definitely should chart are the x-intercept(where y=0) and the y-intercept(where x=0).

Posted
f(x)=2x^2-3 is y=2x^2-3, dilburt.

 

I was responding to the OP in which an expression was to be used as a function.:doh:

Posted
I was responding to the OP in which an expression was to be used as a function.:doh:

Oh sorry, I though you were picking at me for having the foresight to know what he really meant.

  • 2 weeks later...
Posted

If i were doing it i would say what y = when x=0 y would = -3

then what x is when y = 0. You have to use the quadratic formula

 

x = -b (+/-) Root of b^2 - 4ac all divided by 2a

 

for this particular equation a = 2 b = 0 and c = -3

 

you can work out the maximum/minimum via differentiation but you can get a prety good outline of the graph from working out the y and x values.

 

p.s Remember when you square root in the formula it gives two answers still when b = 0.

Posted

It largely depends on what level of accuracy your looking for: a sketch to show you understand the concepts or a plotting to actually be useful? Martianxx's and Ghstofmaxwll's approaches are useful for each occasion respectively.

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