Answer:
A: Exponential
B: Quadratic
C: Linear
Step-by-step explanation:
Let analyse each table
X Y ΔY/ΔX
1 5 xxxxxx
2 11 11 - 5 = 6
3 29 29 - 11 = 18
4 83 83 - 29 = 54
.....................
We can see that, [ ΔY/ΔX ] increases at the fastest rate out of all functions, so it must be exponential function
X Y ΔY/ΔX Δ^2Y/ΔX^2
0 5 xxxxxx xxx
1 7 7 - 5 = 2 xxx
2 13 13 - 7 = 6 6-2 = 4
3 23 23 - 13 = 10 10-6 = 4
..............................
It is a quadratic because Δ^2Y/ΔX^2 = 4 and are constant for all values
X Y ΔY/ΔX
3 20 xxxxxx
6 18 2 / 3
9 16 2 / 3
15 12 2 / 3
..............
It is a linear because ΔY/ΔX = 2/3 are constant for all values
Y = -2x + 5 is parallel to line as the slopes ( = -2) are the same.
the other 2 lines are nether parallel or perpendicular
Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
