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

1. ilikephysics2 Group Title

quadratic equation, do you know the formula?

2. karinewoods17 Group Title

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 Group Title

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

4. karinewoods17 Group Title

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

5. ilikephysics2 Group Title

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 Group Title

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 Group Title

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

8. amistre64 Group Title

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

9. karinewoods17 Group Title

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

10. ilikephysics2 Group Title

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

11. amistre64 Group Title
12. ilikephysics2 Group Title

why are you both copying and pasting?

13. karinewoods17 Group Title

I am not allowed to have a decimal answer though.

14. ilikephysics2 Group Title

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

15. amistre64 Group Title

wolfram does nice graphs :)

16. amistre64 Group Title

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

17. ilikephysics2 Group Title

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

18. karinewoods17 Group Title

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

19. ilikephysics2 Group Title

Then make that happen

20. amistre64 Group Title

"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 Group Title

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 Group Title

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

23. amistre64 Group Title

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

24. amistre64 Group Title

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

25. karinewoods17 Group Title

Whats all that?! lol

26. amistre64 Group Title

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 Group Title

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

28. karinewoods17 Group Title

True

29. karinewoods17 Group Title

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