Question

x^2 + x – 4 = 0 How would this be solved by graphing?

Answer 1

quadratic equation, do you know the formula?

Answer 2

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.

Answer 3

I did that stuff years ago, hang on I will look it up

Answer 4

Thanks. lol I just have trouble with solving the equation by graphing. :/

Answer 5

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

Answer 6

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

Answer 7

The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = -b/2a

Answer 8

solved by graphing means that we graph the equation and look at where it crosses the x axis

Answer 9

\[x = \frac{ -1 \pm \sqrt{1^2 - 4 (1) (-4)} }{ 2 (1) }\]
I get \[x = \frac{ -1 \pm \sqrt{17} }{ 2 }\]
How do I simplify that?

Answer 10

you can leave it in those terms if you want, but type it in the calculator

Answer 11

http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4

Answer 12

why are you both copying and pasting?

Answer 13

I am not allowed to have a decimal answer though.

Answer 14

haha you give her wolfram , nice source and your a moderator...

Answer 15

wolfram does nice graphs :)

Answer 16

to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method

Answer 17

its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers

Answer 18

Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.

Answer 19

Then make that happen

Answer 20

"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 :)

Answer 21

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.

Answer 22

you will get them and understand them, they are not bad at all

Answer 23

completing the square is nice; its the "proof" to the quadratic equation.

Answer 24

\[ax^2+bx+c=0\]
\[x^2+\frac bax+\frac ca=0\]
\[x^2+\frac bax=-\frac ca\]
\[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\]
\[(x+\frac b{2a})^2=\frac{b^2-4ac}{4a^2}\]
\[x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=-\frac b{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 25

Whats all that?! lol

Answer 26

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.

Answer 27

since the completeing the square and the quadratic formula are 2 sides of the same coin.

Answer 28

True

Answer 29

How do I write x^2 + x - 4 = 0 in standard form?