When we want to find the roots of a one-variable function, we look for where its graph intersects the x-axis. In this case, the graph intersects the x-axis at 
The vertex of a parabola is the highest or lowest point on it, depending on whether the leading coefficient of the quadratic function is negative or positive. In this case, we see that the lowest point is 
For the y-intercept, just look for where the graph intersects the y-axis; in this case, that point is 
Using this information, the vertex-form equation of the parabola is
so the factors are two copies of
In this case, the value of
in the equation
was conveniently 1; if that's not the case, you'll want to plug in
to solve for the value of a that gives the correct y-intercept.
Does that help clear things up?
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Solve the initial value problem:
dy——— = 2xy², y = 2, when x = – 1. dxSeparate the variables in the equation above:

Integrate both sides:


Take the reciprocal of both sides, and then you have

In order to find the value of
C₁ , just plug in the equation above those known values for
x and
y, then solve it for
C₁:
y = 2, when
x = – 1. So,


Substitute that for
C₁ into (i), and you have

So
y(– 2) is

I hope this helps. =)
Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>
Answer:
It represents 3x20 +3x15
Step-by-step explanation:
I think it would have to be 1.46.
Explanation: if you subtract 3.56 from 2.1 you’ll get 1.46