Frist 5 in / 1 over 2 = 2.5
So yes she can do that
I dont know if this is what you are looking for but i think this is the answer of the question you are looking for which is the general equation of a horizontal hyperbola:
<span>(x-h)²/a² - (y-k)²/b² = 1 </span>
<span>with </span>
<span>center (h,k)
</span>What you need to do is replace the data and do the calculations
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
The vertex form of all expressions is given below.
We have given that the expressions
We have to write the function in vertex form.
<h3>What is the vertex form of the equation?</h3>
The vertex form of a quadratic function is given by f (x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.
Therefore the first equation
it can be written as
The second equation can be written as
vertex for is,
The third equation is,
Vertex form is,
Forth equation is,
Vertex form is,
To learn more about the vertex form visit:
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