## karinewoods17 3 years ago x^2 + x – 4 = 0 How would this be solved by graphing?

1. ilikephysics2

quadratic equation, do you know the formula?

2. karinewoods17

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. ilikephysics2

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

4. karinewoods17

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

5. ilikephysics2

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. ilikephysics2

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. ilikephysics2

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. karinewoods17

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

10. ilikephysics2

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

11. amistre64
12. ilikephysics2

why are you both copying and pasting?

13. karinewoods17

I am not allowed to have a decimal answer though.

14. ilikephysics2

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. ilikephysics2

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

18. karinewoods17

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

19. ilikephysics2

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. karinewoods17

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. ilikephysics2

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. karinewoods17

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. karinewoods17

True

29. karinewoods17

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