Find an equation for the graph
1 answer:
Answer:
y=6^x+0.612 - 4
Step-by-step explanation:
x = -4 ---> horizontal asymptote
m = 6 ---> use points (0, -1) and (1, 5)
Parent function of the graph:
Our equation:
Add alterations:
- Reflections = N/A
- Vertical & horizontal shifts = down 4
- Vertical & horizontal stretches = left approx 0.612
Final equation: y=6^x+0.612 - 4
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Answer:
(a) On a coordinate plane, a curve starts in quadrant 4 and curves up into quadrant 1
Step-by-step explanation:
A log function does not have a graph that is a line, parabola, or hyperbola. Its graph has a vertical asymptote at x=0, and a x-intercept at x=1. It has positive slope everywhere.
Answer:
The answer is below
Step-by-step explanation:
Plotting the following constraints using the online geogebra graphing tool:
x + 3y ≤ 9 (1)
5x + 2y ≤ 20 (2)
x≥1 and y≥2 (3)
From the graph plot, the solution to the constraint is A(1, 2), B(1, 2.67) and C(3, 2).
We need to minimize the objective function C = 5x + 3y. Therefore:
At point A(1, 2): C = 5(1) + 3(2) = 11
At point B(1, 2.67): C = 5(1) + 3(2.67) = 13
At point C(3, 2): C = 5(3) + 3(2) = 21
Therefore the minimum value of the objective function C = 5x + 3y is at point A(1, 2) which gives a minimum value of 11.
X^2 + y^2 - x - 2*y = 0
To find both coordinates and radius we need to make this equation in circle form:
(x-a)^2 + (y-b)^2 = r^2
x^2 - 2*1/2*x + 1/4 - 1/4 + y^2 - 2*1*y + 1 - 1 = 0
Here we are adding and subtracting numbers in order to get square binomial.
(x - 1/2)^2 + (y-1)^2 = 5/4
coordinates of center are (1/2,1) and the radius is √
To find both coordinates and radius we need to make this equation in circle form:
(x-a)^2 + (y-b)^2 = r^2
x^2 - 2*1/2*x + 1/4 - 1/4 + y^2 - 2*1*y + 1 - 1 = 0
Here we are adding and subtracting numbers in order to get square binomial.
(x - 1/2)^2 + (y-1)^2 = 5/4
coordinates of center are (1/2,1) and the radius is √