## anonymous 3 years ago x^2 + x – 4 = 0 How would this be solved by graphing?

1. anonymous

quadratic equation, do you know the formula?

2. 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.

3. anonymous

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

4. anonymous

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

5. 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

6. 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

7. anonymous

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

8. amistre64

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

9. 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?

10. anonymous

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

11. amistre64
12. anonymous

why are you both copying and pasting?

13. anonymous

I am not allowed to have a decimal answer though.

14. anonymous

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

15. amistre64

wolfram does nice graphs :)

16. amistre64

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

17. anonymous

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

18. anonymous

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

19. anonymous

Then make that happen

20. 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 :)

21. 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.

22. anonymous

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

23. amistre64

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

24. amistre64

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

25. anonymous

Whats all that?! lol

26. 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.

27. amistre64

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

28. anonymous

True

29. anonymous

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