That is in standard form.
This graph has a horizontal asymptote so it is an exponential graph. It also passes through two points (0,-2) and (1,3). The horizontal asymptote is at y=-3.
The unchanged exponential equation is y=a(b)^x +k
For exponential equations, k is always equal to the horizontal asymptote, so k=-3.
You can check this with the ordered pair (0,-2). After that plug in the other ordered pair, (1,3).
This gives you 3=a(b)^1 or 3=ab. If you know the base the answer is simple as you just solve for a.
If you don't know the base at this point you have to sort of guess. For example, let's say both a and b are whole numbers. In that case b would have to be 3, as it can't be 1 since then the answer never changes, and a is 1. Then choose an x-value and not exact corresponding y-value. In this case x=-1 and y= a bit less than -2.75. Plug in the values to your "final" equation of y=(3)^x -3.
So -2.75=(3^-1)-3.
3^-1 is 1/3, 1/3-3 is -8/3 or -2.6667 which is pretty close to -2.75. So we can say the final equation is y=3^x -3.
Hope this helps! It's a lot easier to solve problems like these given either more points which you can use system of equations with, or with a given base or slope.
Given:
Total number of calculators in a box = 10
Defective calculators in the box = 1
To find:
The number of ways in which four calculators be selected and one of the four calculator is defective.
Solution:
We have,
Total calculators = 10
Defective calculators = 1
Then, Non-defective calculator = 10-1 = 9
Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.
Therefore, the four calculators can be selected in 84 ways.
Answer:
-2
Step-by-step explanation:
Start with the vertex form of the equation of a parabola: y - k = a(x - h)^2
Here h = -2, k = -3, x = -1, y = -5. Find a:
-5 - [-3] = a(-1 - [-2])^2, or
-5 + 3 = a(1)^2, or
-2 = a
The unknown coefficient is -2.
Dg stands for "differential geometry"
hope this helps
