Question

Identify any intercepts and test for symmetry for..

Answer 1

\[y=-x ^{2}+3x\]

Answer 2

I no that when you look for x-intercept y=0

Answer 3

and i no that when you look for y-intercept x=0

Answer 4

But the x-intercept in this problem is when i get stuck

Answer 5

y-intercept
x=0
\[y=-0^{2} +3(0)\]
y=0
(0,0)

Answer 6

good, and in this case the y-int is one of the x-ints-- that happens anytime the polynomial does not have a constant term:

x-ints: let y=o and solve for x

-x^2+3x=0
-x(x-3)=0 [factor out a -x]
-x=0 or x-3=0 [set each factor =0]
x=0 or x=3 [solve]
these are the x-ints and the y-int, as ordered pairs (0,0) and (3,0)

Answer 7

wait so x-inter has two points..right?

Answer 8

to test for symmetry you replace x with -x; if you can simplify back to the original rule of the function, then you have symmetry wrt the y-axis (called an even function); if you can simplify to the negative of the original function, then you have symmetry wrt the origin; a function can't have symmetry wrt the x-axis.

this function does not have any of the above symmetry, e.g.,

f(-x)=-(-x)^2+3(-x)=-x^2-3x with is neither f(x0 not -f(x)

Answer 9

(0,0) is an x-int and the y-int
(3,0) is the other x-int

i'll sketch the graph for you next post

Answer 10

ok

Answer 11

Do you think we could exchange emails just incase I have a question and your not on here? :-(

Answer 12

I'll put up my school email for a minute, then when you contact me, i'll give my personal one then, otherwise you know what might happen :})

when i see your icon show i'll post it

Answer 13

ok

Answer 14

get it?

Answer 15

delete it

Answer 16

yea i got it

Answer 17

thanks

Answer 18

was it jesuittampa.org?

Answer 19

you know in reality you'll probably get faster results just coming here, but I'll help if I can; i come on 4-5 night a weeks up to about this time

yes with the two t's

Answer 20

any other questions tonight?

Answer 21

yea two more, should i just keep posting them on this page?

Answer 22

one more is ok; when the page gets too long it bogs down, but we're still ok

Answer 23

Given the two points (6,-2) and (-4,4) find the following : An equation of the circle whose diameter is the line between the two points

Answer 24

The midpoint of the segment is the center of the circle\[x _{m}=\frac{6-4}{2}=1; \ \ y _{m}=\frac{-2+4}{2}=1\]The center of the circle is (1,1).

The distance from the center to either of the two given points is the radius\[r=\sqrt{(6-1)^2+(-2-1)^2}=\sqrt{25+9}=\sqrt{34}\]

The standard form of a circle is \[(x-h)^2+(y-k)^2=r^2\]where (h,k) is the center and r is the radius; so we have h=1, k=1\[(x-1)^2+(y-1)^2=34\]

Answer 25

This page can take one more question

Answer 26

That is it

Answer 27

I wanna thank you though

Answer 28

ok, i'm going to log off then. when i hear from you i'll give my home email. until then :{)

Answer 29

i like to help people that want to try to understand; i get bored just answering questions, and I do this specifically to volunteer-- to give a little back. so it was my pleasure to help

Answer 30

aww thats soo sweet, we need more people like you out here :-), ill be emailing you shortly

Answer 31

g'night

Answer 32

night