1. ## Implied Domain

I have a test tomorrow at 10 am.. it's 9:30 right now. I really need help with understanding implied domain. my teacher never fully explained it, and now i'm panicing. He would like us to put it in interval notation, with teh sideways 8 looking things. PLEASE HELP!

also, how to put a quadratic function into vertex form...
i.e. f(x)= x^2+5x+6 and g(x)= x^2+3

2. Originally Posted by Mmarie
I have a test tomorrow at 10 am.. it's 9:30 right now. I really need help with understanding implied domain. my teacher never fully explained it, and now i'm panicing. He would like us to put it in interval notation, with teh sideways 8 looking things. PLEASE HELP!

also, how to put a quadratic function into vertex form...
i.e. f(x)= x^2+5x+6 and g(x)= x^2+3
for f(x), you must complete the square. (can you continue?)

g(x) is already in it's vertex form. unless you want to get technical and write it as g(x) = (x - 0)^2 + 3

3. Thank you!

well, this is an example and i'm totally confused as to wear my teacher got a 2... you'll understand when you see this:

Finding the vertex by completing the square

h(x)=x^2+9x+20
=x^2+9x+(9/2)^2-(9/2)^2+20
=(x+9/2)^2-81/4+20
=(x+9/2)^2-81/4+80/4
=(x+9/2)^2-1/4
and it says the vertex is (-9/2, -1/4)

..i don't understand where the two that divides the 9 comes from. i'm totally lost!

4. Originally Posted by Mmarie
Thank you!

well, this is an example and i'm totally confused as to wear my teacher got a 2... you'll understand when you see this:

Finding the vertex by completing the square

h(x)=x^2+9x+20
=x^2+9x+(9/2)^2-(9/2)^2+20
=(x+9/2)^2-81/4+20
=(x+9/2)^2-81/4+80/4
=(x+9/2)^2-1/4
and it says the vertex is (-9/2, -1/4)

..i don't understand where the two that divides the 9 comes from. i'm totally lost!
it is a property of perfect squares that the lone constant is always the square of 1/2 the coefficient of x. since the coefficient of x is 9, to have a complete square, we need the constant to be (9/2)^2. since it was not there, we add it, but since we added something that was not there, we subtract it again right after. then we can transform the first three terms into a complete sqaure, and simplify the left overs to get what you see

5. okay yes, i sort of understand. so what about a problem like..

x^2+6x-7 and x^2+5x+6

6. i just tried the first one.. x^2+6x+7

the correct vertex is (-3,-16) but i got (-3,-2)

7. Originally Posted by Mmarie
i just tried the first one.. x^2+6x+7

the correct vertex is (-3,-16) but i got (-3,-2)
you posted x^2 + 6x - 7 last time. which is it?

8. it's +7.. sorry for the confusion!

9. Originally Posted by Mmarie
it's +7.. sorry for the confusion!
it is not (-3,-16)

$\displaystyle y = x^2 + 6x + 7$

$\displaystyle \Rightarrow y = x^2 + 6x + 3^2 - 3^2 + 7$

$\displaystyle \Rightarrow y = (x + 3)^2 - 9 + 7$

$\displaystyle \Rightarrow y = (x + 3)^2 - 2$

thus, the vertex is $\displaystyle (-3,-2)$

10. also, is there anyway you can help me with three more things..

the inverse of--

f(x)=(x+2)/(x+9) and f(x)=(5)/(11x-6)

evaluating functions..
k(x)=3x^2-7 for k(2x^5)

and finding the slope-intercept form when being given a point and the slope.
for example.. m=-3 with the point (-1,2)

PS. i really appreciate all of your help!

11. Originally Posted by Mmarie
also, is there anyway you can help me with three more things..

the inverse of--

f(x)=(x+2)/(x+9) and f(x)=(5)/(11x-6)
say for the first.

f(x) = (x + 2)/(x + 9)

Let y = f(x)

so we have

y = (x + 2)/(x + 9)

for the inverse, switch x and y and solve for y. that will give you the inverse function

that is, solve x = (y + 2)/(y + 9) for y

the other is done in the same way

evaluating functions..
k(x)=3x^2-7 for k(2x^5)

k(2x^5) means, you replace all the x's in k(x) with 2x^5

so, k(2x^5) = 3(2x^5)^2 - 7

now simplify, bearing in mind the laws of exponents

and finding the slope-intercept form when being given a point and the slope.
for example.. m=-3 with the point (-1,2)
the point slope form is: $\displaystyle y - y_1 = m(x - x_1)$. where $\displaystyle m$ is the slope, and $\displaystyle (x_1,y_1)$ is a point the line passes through. if we expand the right hand side of this and solve for y, we get the slope intercept form: y = mx + b
so, you want to solve $\displaystyle y - 2 = -3(x + 1)$ for y