Is anyone here familiar with exponent graphing?

12. Range = -∞ < x  <+∞; Domain = y > 0

13. Range = -∞ < x  <+∞; Domain = y > 3

Step-by-step explanation:

12. f(x) = ½(6)^x

(a) Create a table containing a few values of x and y

x                                                                                  y

-2     ½(6)⁻² = ½(⅙)² = ½ × ¹/₃₆ = ¹/₇₂ = 0.01

-1     ½(6)⁻¹ = ½(⅙)   = ½ × ¹/₁₂ = ¹/₂₄ = 0.08

0     ½(6)⁰  = ½ ×  0 = ½                  = 0.5

 1     ½(6)¹   = ½ ×  6                         =  3

2     ½(6)²  = ½ × 36                        = 18

(b) Draw your axes

Let the x-axis run from -5 to +5 and the y-axis from 0 to 20

(c) Plot your points

They should look like Fig. 1 below.

(d) Draw the graph

Draw a smooth line through the points.

Extend the line in both directions to the edges of the graph.

Your graph should look like Fig. 2.

(e) Identify the domain and range of the function

The domain is the set of all possible x-values that will give real values for y.

It looks like x can vary from -∞ to +∞ and give a real value for y.

The domain is -∞ < x  <+∞.

The range of a function is the spread of all possible y-values.

It looks like y ⟶ ∞ as x ⟶ ∞, but y ⟶ 0 as x ⟶ -∞.

The x-axis is an asymptote. The function can get as close as possible to the x-axis, but it can never be zero or negative.

The range is y > 0.

12. f(x) = 2^x +3

(a) Create a table containing a few values of x and y

x                                                                y    

-4     2⁻⁴ + 3 = (½)⁴ + 3 = ¹/₁₆ + 3 = 3¹/₁₆  =    3.1

-2     2⁻² + 3 = (½)² + 3 =   ¼   + 3 = 3¼ =   3.2

0     2⁰ + 3   =  1    + 3                            =   4

2     2² + 3   =  4   + 3                            =   7

4     2⁴ + 3   = 16   + 3                            = 19

(b) Draw your axes

(c) Plot your points

(d) Draw the graph

Your graph should look like Fig. 3.

(e) Identify the domain and range of the function

The domain is -∞ < x  <+∞.

It looks like y ⟶ ∞ as x ⟶ ∞, but y ⟶ 3 as x ⟶ -∞. The line y = 3 is an asymptote.

The range is y > 3.

Step-by-step explanation:

h = 8 cm,  r = 3 cm

A = 2×πr² + 2πr×h = 2πr×(r + h)

A = 2π×3×(3 + 8) =  66π cm² ≈ 207.35 cm²

h = 2×8 = 16 cm ,      r = 3 cm ,   2×66π = 132π cm²

A = 2π×3×(3 + 16) =  114π cm² ≠ 132π cm²  

h = 8 cm ,        r = 2×3 = 6 cm  ,   2×66π = 132π cm²

A = 2π×6×(6 + 8) =  168π cm² ≠ 132π cm²