anonymous
  • anonymous
x^2 + x – 4 = 0 How would this be solved by graphing?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
quadratic equation, do you know the formula?
anonymous
  • anonymous
yes. \[x =\frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}\] But I need the vertex, domain and range, x and y intercepts, and the axis of symmetry.
anonymous
  • anonymous
I did that stuff years ago, hang on I will look it up

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anonymous
  • anonymous
Thanks. lol I just have trouble with solving the equation by graphing. :/
anonymous
  • anonymous
the x and y intercepts are going to be the equation factored out in the quadratic formula...I'm sure you can do that right
anonymous
  • anonymous
the domain will be all x values not equal to the x and y intercepts that you get, and they can not make the denominator 0
anonymous
  • anonymous
The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = -b/2a
amistre64
  • amistre64
solved by graphing means that we graph the equation and look at where it crosses the x axis
anonymous
  • anonymous
\[x = \frac{ -1 \pm \sqrt{1^2 - 4 (1) (-4)} }{ 2 (1) }\] I get \[x = \frac{ -1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?
anonymous
  • anonymous
you can leave it in those terms if you want, but type it in the calculator
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4
anonymous
  • anonymous
why are you both copying and pasting?
anonymous
  • anonymous
I am not allowed to have a decimal answer though.
anonymous
  • anonymous
haha you give her wolfram , nice source and your a moderator...
amistre64
  • amistre64
wolfram does nice graphs :)
amistre64
  • amistre64
to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method
anonymous
  • anonymous
its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers
anonymous
  • anonymous
Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.
anonymous
  • anonymous
Then make that happen
amistre64
  • amistre64
"solve by graphing" is prolly the least exact method of finding a solution. But it is a viable in order to see if you ar in the right ballpark. The wolf does go ahead and spit out al the properties such as the exact figures :)
anonymous
  • anonymous
Yeah its just apart of learning the different methods to solving quadratic equations. I also have to find the answer using factoring, quadratic formula, and completing the square.
anonymous
  • anonymous
you will get them and understand them, they are not bad at all
amistre64
  • amistre64
completing the square is nice; its the "proof" to the quadratic equation.
amistre64
  • amistre64
\[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=-\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\] \[(x+\frac b{2a})^2=\frac{b^2-4ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=-\frac b{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
Whats all that?! lol
amistre64
  • amistre64
thats the a general quadratic setup, and by running thru the "complete the square" process, we end up with the quadratic formula. completeing the square is the proof to the quadratic formula.
amistre64
  • amistre64
since the completeing the square and the quadratic formula are 2 sides of the same coin.
anonymous
  • anonymous
True
anonymous
  • anonymous
How do I write x^2 + x - 4 = 0 in standard form?

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