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graph the function g(x) = x3 − x2 − 4x + 4. (an actual graph that you can attach)

see below

Step-by-step explanation:

g(x) = x^3 − x^2 − 4x + 4

We know the graph will have up to 3 zero's because it is a cubic

g(x) = x^3 − x^2 − 4^x + 4

I will factor by grouping, taking an x^2 from the first 2 terms and -4 from the last 2 terms

  g(x)=  x^2(x-1) -4(x-1)

Now factor out x-1

  g(x)= (x-1)(x^2 -4)

We can factor the (x^2-4) as a difference of squares

g(x) = (x-1) (x-2)(x+2)

Using the zero product property

0= (x-1) (x-2)(x+2)

x-1 =0   x-2 =0   x+2=0

We have zeros at x=1   x=2  and x=-2

Then we can plot points to determine where the function is between the points We will pick negative infinity 0 1.5 and infinity

at g(-inf) =  -inf  because x^3 dominates and that goes to -infinity

at g(0) = 0+000+4 =4

at g(1.5) =-.875

at  g(inf)=because x^3 dominates and that goes to infinity

answer are

step-by-step explanation:

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!