Question

Hello, so I need to do revisions and I don't understand Linear Functions well. How do I find the four sequence of the expression n (n-1)-5, I also need help with, What function rule best represents the data in the table.
x |-3, -2, -1, 0, 1
y| -17, -14, -11, -8, -5
The last thing I need help with is What quadratic rule represents the data in the table.
x| -1, 0, 1, 2, 3
y| 6, 5, 6, 9, 14

Thanks for helping me:)

Answer 1

n , n-1 , -5
well maybe
n - (n-1) = (n-1) - (-5)
1 = n -1 +5 = n + 4
n = -3
so
-3, -4, -5, -6, -7, -8 ...etc

Answer 2

when x goes up by 1 (for example from 0 to 1)
y goes up by 3 (for example from -8 to -5)
so the slope is 3
form y = 3 x + b
what is b?
put any point in
(1,-5)
-5 = 3(1) +b
b = -8
so y = 3 x - 8

now check with another point like (-3, -17)
if x = -3
y = 3(-3) -8
y = -9 - 8
y = -17 whew, sure enough

Answer 3

I your function definitiion is n(n-1)-5
if n=1 ----> 1(0) - 5 = -5
if n=2 ----> 2(1) - 5 = -3
if n=3 ----> 3(2) - 5 = 0
if n=4 ......

for the 2st, if you notice that for each increase of 1 in the x, you have an increase of 3 in the y, so we can start by writing it as
y = 3x + b
If you look at the x=0, y = -8, we see that the y-intercept is -8, so b = -8

your function is y = 3x - 8

for the 3rd, notice that for each increase of x of 1, we do not have the same
increase in y,
but if we take the increase in the previous increases, we do get a constant, so you
do have a quadratic. (we could have just taken their word for it, but this is more fun)

so use (-1,6), (0,5) and (1,6)
(I already have a sneaky suspicion that my vertex will be (0,5) )

let f(x) = ax^2 + bx + c
for (-1,6) ---> a - b + c = 6
for (0,5) ----> 0 + 0 + c = 5 , so c = 5
for (1,6) ----> a + b + c = 6
add the first and last: 2a + 10 = 12, a = 1
in a+b+c = 6
1+b+5=6
b = 0

f(x) = x^2 + 5

Answer 4

x |-3, -2, -1, 0, 1
y| -17, -14, -11, -8, -5
note that y increases by 3 when x increases by 1
So, start with y = 3x
But, y(0) = -8, so adjust it to y = 3x-8

x| -1, 0, 1, 2, 3
y| 6, 5, 6, 9, 14
Here, note that the amount of change in y (∆y) for each bump in x does not remain the same. So, look at the differences between y-values
y| 6, 5, 6, 9, 14
∆1| -1, 1, 3, 5
Now look at the 2nd differences:
∆2: 2, 2, 2
A quadratic relationship will have constant 2nd differences.
A linear relationship will have constant 1st differences (slope)
Now, note that the y-values are symmetric about x=0, and y(0)=5
So, start with y = ax^2 + 5
Now pick another point to find the value of a.
y(1) = 6 so
a*1^2+5 = 6
a+5 = 6
a = 1
So, y = x^2 + 5

Answer 5

x| -1, 0, 1, 2, 3
y| 6, 5, 6, 9, 14

y = a x^2 + bx + c
well to make it easy y = c when x = 0
so
c = 5
so
y = a x^2 + b x + 5
put in (-1, 6) for example
6 = a - b + 5 so a-b = 1
now another say (+1, 6)
6 = a + b + 5 so a + b = 1
add equations and get 2 a = 2 so a = 1
then b = 0 ( which if we were smart we would already know from symmetry)
so
y = x^2 +5

Answer 6

Thanks so much, everyone:)