Answer:
B) The maximum y-value of f(x) approaches 2
C) g(x) has the largest possible y-value
Step-by-step explanation:
f(x)=-5^x+2
f(x) is an exponential function.
Lim x→∞ f(x) = Lim x→∞ (-5^x+2) = -5^(∞)+2 = -∞+2→ Lim x→∞ f(x) = -∞
Lim x→ -∞ f(x) = Lim x→ -∞ (-5^x+2) = -5^(-∞)+2 = -1/5^∞+2 = -1/∞+2 = 0+2→
Lim x→ -∞ f(x) = 2
Then the maximun y-value of f(x) approaches 2
g(x)=-5x^2+2
g(x) is a quadratic function. The graph is a parabola
g(x)=ax^2+bx+c
a=-5<0, the parabola opens downward and has a maximum value at
x=-b/(2a)
b=0
c=2
x=-0/2(-5)
x=0/10
x=0
The maximum value is at x=0:
g(0)=-5(0)^2+2=-5(0)+2=0+2→g(0)=2
The maximum value of g(x) is 2
Answer:
1/3
Step-by-step explanation:
<em>Method 1.</em>
slope = rise/run
Rise is vertical distance.
Run is horizontal distance.
Find two points that are easy to read (on grid intersections):
(2, -1) and (5, 0).
Start at (2, 1). You need to go to (5, 0) by moving only vertically and horizontally. Go up 1 unit. That is a rise of 1. Now go right 3 units. That is a run of 3.
rise = 1
run = 3
slope = rise/run = 1/3
<em>Method 2.</em>
Use the slope formula and two points on the line.
Use points (2, -1) and (5, 0).
slope = 1/3
To find if a series is either geometric or arithmetic:
it must satisfy this property:
Arithmetic:
a(n+1) - a(n) = const
Geometric:
a(n+1)/a(n) = const
In your case:
r1 = 7 -4 = 3
r2 = 12 - 7 = 5
r1 != r2 (not arirthmetic)
Geometric check:
r1 = 7/4
r2 = 12/7
r1 != r2 (not Geometric)
so neither.
