Cheyenne multiplied three numbers together and divided by 4. The result was a negative value. Which of the following statements
1 answer:
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Answer:
a = -0.32
b = 16
c = - 168
Step-by-step explanation:
In the figure attached, the quadratic function is shown. We can model it as follows:
y = a*(x - x1)*(x - x2)
where (x1, 0) and (x2, 0) are its roots
From the picture, roots are (15, 0) and (35, 0), replacing them into the equation:
y = a*(x - 15)*(x - 35)
and the vertex is at (25,32), replacing into the equation:
32 = a*(25 - 15)*(25 - 35)
32 = a*(-100)
a = 32/(-100)
a = -0.32
Applying distributive property:
y = -0.32*(x - 15)*(x - 35)
y = -0.32*(x² - 35x - 15x + 525)
y = -0.32x² + 16x - 168
which corresponds to the general form:
y = ax² + bx + c
Complete Question:
The complete question is shown on the first uploaded image
Answer:
a) The graph of f(y) versus y. is shown on the second uploaded image
b) The critical point is at y = 0 and the solution is asymptotically unstable.
c)The phase line is shown on the third uploaded image
d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image
Step-by-step explanation:
Step One: Sketch The Graph of f(y) versus y
Looking at the given differential equation
for -∞ <
< ∞
We can say let = f(y) =
Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.
Step Two : Determine the critical point
To fin the critical point we have to set = 0
This means = 0
For this to be possible = 1
which means that =
which implies that y = 0
Hence the critical point occurs at y = 0
meaning that the equilibrium solution is y = 0
As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.
Step Three : Draw the Phase lines
A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.
In each of the intervals bounded by the equilibrium draw an upward
pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.
This phase line would solely depend on y does not matter what t is
On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).
For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper
Step Four : Draw a Solution Curve
A solution curve is a curve that shows the solution of a DE (deferential equation)
Here the solution curve would be drawn on the ty-plane
So the t-axis(x-axis) is its the equilibrium that is it is the solution
If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)
