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kap26 [50]
3 years ago
10

What can you say about the y-values of the two functions f(x)=-5^x +2 and g(x)=-5x^2+2?

Mathematics
1 answer:
Kryger [21]3 years ago
7 0

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

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3 years ago
Construct a linear equation for the table. What is the slope of the line
Klio2033 [76]

We have points (-3, 4), (-1, 2), (4, -3) and (6,-5)

Let's verify it's a line by calculating the slopes between successive points.

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3 0
2 years ago
Write 240 and 1500 as products of their prime factor
Marina CMI [18]
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4 0
3 years ago
Read 2 more answers
HELP!! Algebra help!! Will give stars thank u so much &lt;333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
2 years ago
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