anonymous
  • anonymous
i forgot how to graph when you get the answers from the quadratic formula can someone help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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OakTree
  • OakTree
Sure! So what's the quadratic you have?
anonymous
  • anonymous
3x^2-2x-5 the answers are 5/3 and -1
OakTree
  • OakTree
That looks right. So the first thing we want to notice is the fact that the first term is positive. What can you tell me about the graph based on that?

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anonymous
  • anonymous
it should be opening up?
OakTree
  • OakTree
Right. So now we look at the roots. Do you know what the roots look like on a graph?
anonymous
  • anonymous
i just can not remember for the life of me how to graph it like the x and the y
OakTree
  • OakTree
Okay, that's fine. So the roots actually mean the points where the graph crosses the x-axis. Do you want me to explain why?
anonymous
  • anonymous
yes please
OakTree
  • OakTree
Okay. So when you plug a root into a function, what do you get? So if the quadratic is \[f(x) = x^2 + x + 6\]The roots are 3 and -2. So what do you get when you plug those in?
anonymous
  • anonymous
0
anonymous
  • anonymous
you get 0 when you put in -2 and for 3 you get 18
OakTree
  • OakTree
The idea is you get 0. So basically, when you plug a root in, you get the answer\[f(x) = 0\]But f(x) is the same as y, so when you plug in a root as the x, you get 0 as the y. In essence, this is an ordered pair - where the root is the x and 0 is always the y. But all pairs where y=0 are on the x axis. So the roots are just always where the graph touches the x-axis. Do you get it?
anonymous
  • anonymous
Yes! lol thank you! I just needed something to trigger my memory! So for x which ever one you plug in and it allows f(x)=0 then is the number you use to graph?
OakTree
  • OakTree
Kind of. But I'm glad you get the explanation! To graph, we just need to put points where the roots are and find the vertex and connect them. Do you remember how to find the vertex of a parabola?
anonymous
  • anonymous
i feel as if it is the square root of something, but that i am not sure of
OakTree
  • OakTree
It's like this: to find the x-value of the vertex, you use the formula\[x = \frac{ -b }{ 2a }\]And then you plug in whatever you get for x into the function to get your y. Get it?
anonymous
  • anonymous
haha wow yeah this summer i forgot all of algebra 2 w/ trig
OakTree
  • OakTree
That's fine. So you can handle the problem now?
anonymous
  • anonymous
i sure can thank you so much
OakTree
  • OakTree
You're very welcome.

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