!HELP! Find The Exponential Equations For The Following Graph Below
2 answers:
You might be interested in
Answer:
second option
Step-by-step explanation:
Given
+ 6x² + 5 = 0
let u = x², then
u² + 6u + 5 = 0 ← in standard form
(u + 1)(u + 5) = 0 ← in factored form
Equate each factor to zero and solve for u
u + 1 = 0 ⇒ u = - 1
u + 5 = 0 ⇒ u = - 5
Change u back into terms of x, that is
x² = - 1 ( take the square root of both sides )
x = ± = ± i ( noting that
= i ), and
x² = - 5 ( take the square root of both sides )
x = ± = ±
= ±
×
= ± i
Solutions are x = ± i and x = ± i
Answer:
2) D: x = [0, 24]
3) R: y = [0, 384]
4) see graph
<u>Step-by-step explanation:</u>
Eric's regular wage is $12 per hour for all hours less than 9 hours.
The minimum number of hours Eric can work each day is 0.
f(x) = 12x for 0 ≤ x < 9
Eric's overtime wage is $18 per hour for 9 hours and greater.
The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).
f(x) = 18(x - 8) + 12(8)
= 18x - 144 + 96
= 18x - 48 for 9 ≤ x ≤ 24
The daily wage where x represents the number of hours worked can be displayed in function format as follows:
2) Domain represents the x-values (number of hours Eric can work).
The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.
D: 0 ≤ x ≤ 24 → D: x = [0, 24]
3) Range represents the y-values (wage Eric will earn).
Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.
The minimum hours he can work in one day is 0.
f(x) = 12x
f(0) = 12(0)
= 0
The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>
f(x) = 18x - 48
f(24) = 18(24) - 48
= 432 - 48
= 384
D: 0 ≤ y ≤ 384 → D: x = [0, 384]
4) see graph.
Notice that there is an open dot at x = 9 for f(x) = 12x
and a closed dot at x = 9 for f(x) = 18x - 48
