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SIZIF [17.4K]
3 years ago
8

An exponential function with a base of 1/2 has been vertically compressed by a factor of 3/4 and reflected in the x-axis. It's a

symptote is the line y= -4. What is the equation of the function?
Mathematics
1 answer:
joja [24]3 years ago
7 0
We are asked in this problem to determine the expression of an exponential function given the data above. The standard form of an exponential function is expressed as y = a^x + b was a is the base constant and b is another constant. In this case, a is given to be equal to 1/2. since the graph is said to be shrunk down to 3/4 then the initial expression would have to be y = 3/4*(1/2)^x + b. b is determined through the given asymptote in the problem. An asymptote is an axis or the line in which the graph approaches but never touches. b is equal to -4. In this case, the overall expression is y = 3/4 *(1/2)^x - 4.
You might be interested in
Please teach me how to solve part b, thank you.​
MA_775_DIABLO [31]

Part (a)

The locus is the set of points that satisfy the given conditions. Namely that the point P needs to be 1 unit from the line given.

Think of a median along a highway. Then imagine the two shoulders of the road. The locus describes the two parallel lines that are to this given median. In other words, the median is y = x+2 and the road shoulders are unknown for now.

Though we do know that parallel lines have equal slopes (but different y intercepts).

Therefore, the answers to the next section will be of the form y = x+k for some real number k.

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

Part (b)

Use your favorite graphing tool to plot y = x+2. I recommend GeoGebra since I use it all the time, and it's what I used to make the drawing below. Desmos is another tool you can use. Both are free.

Note that point A(0,2) is on the line y = x+2. The line perpendicular to y = x+2 and goes through (0,2) is y = -x+2. From here, plot a circle centered at (0,2) with radius 1. The equation of this circle is x^2 + (y-2)^2 = 1

The intersection points of the circle and y = -x+2 will provide starting points, so to speak, for the two shoulder lines I mentioned in part (a).

The approximate locations of those points are:

B = (0.71, 1.29)

C = (-0.71, 2.71)

From here, it shouldn't be too hard to find the y = mx+b forms of the locus lines. Let me know if you need help with that, or if you need me to check your answers.

Side note: Why is it called a locus? I'm not sure, but my gut feeling tells me it has to do with the insects of a similar sounding name (locus the math term and locust the insect). Imagine millions of them forming various shapes in the sky, similar to how many points are used in the 2D plane to create various curves.

5 0
2 years ago
The difference in the x-coordinates of two points is 3, and the difference in the y-coordinates of the two points is 6. What is
nirvana33 [79]

Answer:

2

Step-by-step explanation:

6/3=2

5 0
3 years ago
Read 2 more answers
I NEED THIS NOW DILATIONS HELP ME PLS
docker41 [41]

Answer:

reflect over the y-axis, then translate (x+(-1), y + (1))

5 0
3 years ago
Can somebody explain this in a basic way and help me graph it??!!
7nadin3 [17]

Hello!

To graph this, you will want to create two points based off of the equation. You plug an x-value into the equation, and you put the y-value outcome with it as an ordered pair. You do this twice, giving you two points to connect.

For example, let's plug in an x-value 1, giving us (1,y)

y= 1-3 (We remove the absolute value)

y=-2 (Subtract)

This gives us one point, (1,-2)

Let's try using -2 as another point.

y= 2-3 (Remove absolute value)

y= -1 (Subtract)

This gives us the point (-2,-1)

To graph, you would draw a straight line through the points (1,-2) and (-2,-1). Note you could do this with any x-value and you would get the same line.

I hope this helps!


3 0
2 years ago
Read 2 more answers
In math class help!!!!!
Pani-rosa [81]

i need more than this

8 0
2 years ago
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