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karinewoods17
Group Title
x^2 + x – 4 = 0 How would this be solved by graphing?
 one year ago
 one year ago
karinewoods17 Group Title
x^2 + x – 4 = 0 How would this be solved by graphing?
 one year ago
 one year ago

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ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
quadratic equation, do you know the formula?
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
I did that stuff years ago, hang on I will look it up
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
Thanks. lol I just have trouble with solving the equation by graphing. :/
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = b/2a
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
solved by graphing means that we graph the equation and look at where it crosses the x axis
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
\[x = \frac{ 1 \pm \sqrt{1^2  4 (1) (4)} }{ 2 (1) }\] I get \[x = \frac{ 1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
you can leave it in those terms if you want, but type it in the calculator
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
why are you both copying and pasting?
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
I am not allowed to have a decimal answer though.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
haha you give her wolfram , nice source and your a moderator...
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
wolfram does nice graphs :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
Then make that happen
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
"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 :)
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
you will get them and understand them, they are not bad at all
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
completing the square is nice; its the "proof" to the quadratic equation.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}\frac ca\] \[(x+\frac b{2a})^2=\frac{b^24ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^24ac}}{2a}\] \[x=\frac b{2a}\pm\frac{\sqrt{b^24ac}}{2a}\] \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
Whats all that?! lol
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
since the completeing the square and the quadratic formula are 2 sides of the same coin.
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
True
 one year ago

karinewoods17 Group TitleBest ResponseYou've already chosen the best response.0
How do I write x^2 + x  4 = 0 in standard form?
 one year ago
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