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karinewoods17
x^2 + x – 4 = 0 How would this be solved by graphing?
quadratic equation, do you know the formula?
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.
I did that stuff years ago, hang on I will look it up
Thanks. lol I just have trouble with solving the equation by graphing. :/
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
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
The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = -b/2a
solved by graphing means that we graph the equation and look at where it crosses the x axis
\[x = \frac{ -1 \pm \sqrt{1^2 - 4 (1) (-4)} }{ 2 (1) }\] I get \[x = \frac{ -1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?
you can leave it in those terms if you want, but type it in the calculator
http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4
why are you both copying and pasting?
I am not allowed to have a decimal answer though.
haha you give her wolfram , nice source and your a moderator...
wolfram does nice graphs :)
to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method
its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers
Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.
Then make that happen
"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 :)
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.
you will get them and understand them, they are not bad at all
completing the square is nice; its the "proof" to the quadratic equation.
\[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=-\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\] \[(x+\frac b{2a})^2=\frac{b^2-4ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=-\frac b{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Whats all that?! lol
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.
since the completeing the square and the quadratic formula are 2 sides of the same coin.
How do I write x^2 + x - 4 = 0 in standard form?