y=-3(x-2)^2 +2
In the xy plane, the line l passes through the point (-1,3) and the vertex of the parabola with the equation above. What is the slope of line l?
A)-2/3
B)-1/2
C)-1/3
D)1/3
^I don't know how to do this
================================
I do.
y=-3(x-2)^2 +2
This is a dead giveaway.
if x = 3, (x-2)^2 = +1
if x = 1, (x-2)^2 = +1
In other words this parabola is the same on both sides of x = 2 !!!! :) There is the vertex.
What else do we know that we really do not need?
If at x = 2, y = 2 , so that is the vertex
if x is big + or big -, then y is big - so this parabola sheds water.
line goes through
(2,2) and the given point (-1,3)
what is the slope?
No problem
(y2-y1)/(x2-x1) = (3-2)/(-1-2) = -1/3
y=-3(x-2)^2 +2
In the xy plane, the line l passes through the point (-1,3) and the vertex of the parabola with the equation above. What is the slope of line l?
A)-2/3
B)-1/2
C)-1/3
D)1/3
^I don't know how to do this
Let function f(x) be defined by the equation f(x)=1/2-x
If m is a positive integer then f(1/m) =
A) m/2m-1
B) m/m^2-1
C)1/2-m
D)2-m
^I got A?
The value of y varies with x according to the equation y=a(x-2)(x+1), where a is less than 0. As the value of x increases from 0 to 5 which of the following best describes the behavior of y?
A)it increases and then decreases
B)IT decreases and then increases
C)IT increases only
D)It decreases only
^I don't know. sorry
5 answers
first: find the x,y for the vertex. Hint, what happens when x=2? so now you have two points on the line, what is the slope.
second: f(1/m)=1/2 - 1/m
I dont see (the way you typed them) that any of the answers are correct. But I can see A as correct, if you meant f(x) to be 1/(2-x)
Third: put in some values: x=1/2, then x=0 then 1 then 2 then 4
second: f(1/m)=1/2 - 1/m
I dont see (the way you typed them) that any of the answers are correct. But I can see A as correct, if you meant f(x) to be 1/(2-x)
Third: put in some values: x=1/2, then x=0 then 1 then 2 then 4
Let function f(x) be defined by the equation f(x)=1/2-x
If m is a positive integer then f(1/m) =
A) m/2m-1
B) m/m^2-1
C)1/2-m
D)2-m
^I got A?
==========================
IF you meant what you typed
f(x) = 1/2 - x and not 1/(2-x)
then
f(1/m) = 1/2 - 1/m
= (m-2)/2m
BUT I think you typed it wrong
f(x) = 1/(2-x)
f(1/m) = 1 / (2- 1/m)
= m/(2m - 1) So A but still a typo
NEED PARENTHESES
If m is a positive integer then f(1/m) =
A) m/2m-1
B) m/m^2-1
C)1/2-m
D)2-m
^I got A?
==========================
IF you meant what you typed
f(x) = 1/2 - x and not 1/(2-x)
then
f(1/m) = 1/2 - 1/m
= (m-2)/2m
BUT I think you typed it wrong
f(x) = 1/(2-x)
f(1/m) = 1 / (2- 1/m)
= m/(2m - 1) So A but still a typo
NEED PARENTHESES
The value of y varies with x according to the equation y=a(x-2)(x+1), where a is less than 0. As the value of x increases from 0 to 5 which of the following best describes the behavior of y?
A)it increases and then decreases
B)IT decreases and then increases
C)IT increases only
D)It decreases only
===============================
y=a(x-2)(x+1) LET a=-1
y = -(x-2)(x+1)
parabola
sheds water, big x gives -oo y
zero at x = -1 and x =2, hump between
top of hump halfway between zeros
x = (2+1)/2 + -1 = -1+3/2 = 1/2
goes up from x = 0 to 1/2
then down to x = 2 and continues down forever
A)it increases and then decreases
B)IT decreases and then increases
C)IT increases only
D)It decreases only
===============================
y=a(x-2)(x+1) LET a=-1
y = -(x-2)(x+1)
parabola
sheds water, big x gives -oo y
zero at x = -1 and x =2, hump between
top of hump halfway between zeros
x = (2+1)/2 + -1 = -1+3/2 = 1/2
goes up from x = 0 to 1/2
then down to x = 2 and continues down forever
Yes, it was a typo. Sorry @-@ Thank you.