points of intersection

[denise]

hello. i'm not really sure wat thery are askin. when u are asked to find the points of intersection, are you trying to find the zeros. or what are u trying to find.
heres an example:


find the points of intersection:

f(x) = x^(2) - 4 and g (x) = 3x + 20

 

[soroban]

Hello, denise!

Find the points of intersection:

f(x) = x² - 4 and g (x) = 3x + 20
.

We want the point(s) common to both graphs.

Set the functions equal to each other and solve: . x² - 4 .= .3x + 20 . . . etc.

 

[~Midnight.Kitten~]

do you use the PEMDAS rule??

 

[~Midnight.Kitten~]

PEMDAS, Parenthese,exponents,multipication,division,addition,subtraction[/img][/list]

 

[tkhunny]

Well, if you need it, but that wouldn't be my first guess.

The idea is to solve the equation. Soroban gave you the equation to solve. How do you suppose we shall do that? It looks quadratic to me.

 

[denise]

still need help

would u be so kind to give me an example or something?

 

[tkhunny]

x^2 - 4 = 3x + 20

x^2 - 3x - 24 = 0

Is it looking familiar, yet?

 

[denise]

yeah. i did that. now i factor rite? well thats were i'm stuck. do i now use the pemdas or what do i do?

 

[Denis]

Well, since you don't seem familiar with the quadratic equation,
then check your g(x) = 3x + 20; is it g(x) = 2x + 20 ?

If so, then you get x^2 - 2x - 24 = 0,
easily factored: (x - 6)(x + 4) = 0

And you, Black Cat, you got PEMDAS well defined, but it has
nothing to do with this: quit confusing the issue!

 

[denise]

the equation is g(x) = 3x + 20

 

[Denis]

Well then, you need to solve using the quadratic equation;
if you don't know what that is, ask your teacher: this ain't no classroom.

Or look it up using http://www.google.com