Find the exponential function of the 2 points (3, 7) and (5, 63). Use
the formula f(x) = a(b)”.

We obtained the two exponential functions:

Step-by-step explanation:

As we know that the exponential function is of the form

f(x) = abˣ

Given the points

We know these points belong to the exponential function.

so substituting the values (3, 7) and (5, 63) in the function

putting (3, 7)

y = abˣ  

7 = ab³

also putting (5, 63)

y = abˣ

63 = ab⁵

Considering the 2nd equation

63 = ab⁵

as

so

63 = ab³×b²  

substituting 7 = ab³ in 63 = ab³×b²  

63 = 7 × b²

b² = 63/7

b² = 9

b = ± 3

If b = 3

plug in b = 3 in the equation 7 = ab³ to find the value 'a'

7 = ab³

7 = a(3)³

7 = a × 27

a = 7/27

so, a = 7/27 and b = 3 would give us the function

y = abˣ

if b = -3

plug in b = -3 in the equation 7 = ab³ to find the value 'a'

so, a = -7/27 and b = -3 would give us the function

y = abˣ

Thus, we obtained the two exponential functions:

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!