which point lies on the graph of the function shown below?
y= -x^2 + 4x -2

(-1, 3)
(0,2)
(2, -2)
(1,1)

(1, 1)

Step-by-step explanation:

We need to evaluate the given expression for each of the x-values and check if the y-value agrees with the y-value given in the pairs:

1) First point (-1,3): use x=-1 and check if y results in "3":

we obtained y=-7, so the point (-1,3) is Not on the graph of the function

2) Second point (0,2): use x=0 and check if y results in "2":

we obtained y=-2, so the point (0,2) is Not on the graph of the function

3) Third point (2,-2): use x=2 and check if y results in "-2":

we obtained y=2 (and not -2), so the point (2,-2) is Not on the graph of the function

4) Fourth point (1,1): use x=1 and check if y results in "1":

we obtained y=1 (the same y as in the given coordinate pair), so the point (1,1) IS on the graph of the function

∞ this is your sign

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}