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

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

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

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

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

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

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

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.1
Then make that happen
 2 years 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 :)
 2 years 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.
 2 years 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
 2 years ago

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

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

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

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