# graphing calcluators

## Recommended Posts

i am supposed to find the roots of quadratic equations but they never showed how to do it without a graphing calculator and i can't figure out how to get any of the online ones to tell me the vertex or the roots.

##### Share on other sites

i am supposed to find the roots of quadratic equations but they never showed how to do it without a graphing calculator and i can't figure out how to get any of the online ones to tell me the vertex or the roots.

Haven't you studied the "quadratic formula" ? You should be able to find the roots of a quadratic equation with no calculator at all.

##### Share on other sites

What's the equation - and what have you tried?

I presume you can write it as ax2+bx+c=0 (if b = 0 it's either really easy or impossible to find real roots).

The next step is to see if you can write it any other way - Do you understand how to factorise a quadratic? - And what you can easily tell from a quadratic written in a factorised form?

So let us know where you started and where you have got to and then we can move on.

##### Share on other sites

All written below is for ax^2+bx+c=0

x1=[-b+sqrt(b^2-4ac)]/2a

x2=[-b-sqrt(b^2-4ac)]/2a

If b^2-4ac=0 than the equation has only 1 solution, if b^2-4ac>0 than it has 2 solutions, if b^2-4ac<0 than it has no real solutions.

##### Share on other sites

yeah but i don't know how to use it to find the vertex or the roots

##### Share on other sites

yeah but i don't know how to use it to find the vertex or the roots

Well for the vertex:

First of all lets define the vertex as $(h,k)$

so $h= \frac{-b}{2a}$

and $k= \frac{(4ac-b^{2})}{4a}$

The roots are just the solutions given by:

$x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

All this applies for any quadratic of the form:

$ax^{2}+bx+c=0$

But also notice how h and k could be given by completing the square and therefore changing the equation from standard form to vertex form.

##### Share on other sites

Well for the vertex:

First of all lets define the vertex as $(h,k)$

so $h= \frac{-b}{2a}$

and $k= \frac{(4ac-b^{2})}{4a}$

The roots are just the solutions given by:

$x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

All this applies for any quadratic of the form:

$ax^{2}+bx+c=0$

But also notice how h and k could be given by completing the square and therefore changing the equation from standard form to vertex form.

If you want to sketch the graph, you can use roots as cross-points of graph and x-axis, vertex as minimum or maximum (depends if a<0 [minimum] or a>0 [maximum]) and A(0,y) (as cross-point of graph and y-axis) where y is function's value for x=0 (IOW y=c).

Edited by Djordje

## Create an account

Register a new account