Critical Point Type for a Quadratic Function

Critical point type for a quadratic function

We will use the second derivative test to find a maximum or minimum value for an
equation. By definition, if y'=0, then there must be a local maximum or minimum at
that point. The main problem is that we have more possible situations and have
several derivatives, so we need to think harder about how we will decide. However, it
will involve the second derivatives again.
Let us examine an example in which we have a function that is quadratic.

We will
consider the case where the function is of the form ax2+bxy+cy2.
The equation has a critical point at the origin because if you take the derivative with
respect to x, and if you plug x = y = 0, you'll get 0. You can also see that all of these
numbers are much smaller than x and y when x and y are small, so the linear
approximation, the tangent plane to the graph, is really just w = 0.

So we took a look at something that starts with x squared plus 2xy plus 3y squared.
We can rewrite this as x plus y squared plus 2y squared. Since each of these terms
must be non-negative, the origin will be a minimum. A generalization of this identity
can be obtained by first completing the square in order to obtain, in general, a
quadratic equation. We will assume that a is non-zero so that we may proceed.
Now the square of this looks like it might be the beginning of a binomial expansion.

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This is similar to what we did earlier with the binomial expansion a(x + y)2 = x2 +
2axy + y2. If we put b over 2a times y and square this expression, then the
cross-term 2xy will become b over ax by. Of course, now we also
get some y2 terms
out of this.
How many y squares do we get? We can multiply b squared over 4a to get the
number of y squares, c times y squared. We want to find c minus b squared over 4a.
Let's look at that again. If we expand this thing. we will get ax squared plus b over 2a
times 2xy. That's going to be our bxy. But we also get squared over 4y squared times
a. That's squared over 4a y squared. And that cancels out with this item here. Left
with cy squared.
If we expand (x + 2)2, we get x2 2x(2) squared over 2a squared plus a times x
squared divided by a. Notice that the two a's cancel out, leaving us with bxy over a
squared y squared plus cy over a squared minus b square over 4a y squared. Here,
a the a and the simplify, and now these two terms simplify to give us just c squared
in the end.
So we'll rewrite this equation in slightly different form. ITake 4a to the second
power and add it to x plus b over 2a times y squared. This will give us the same thing
as before, but with the 4a squared term canceled out by another 4a squared term. In
other words, nothing changes numerically; we just clear the denominator on one side
of the equation by adding the same quantity to both sides. Next, subtract b squared y
squared from both sides of the equation.
The equation can be expressed as the sum of two squares, which means that the
origin is a minimum point on the graph. If you have the equation expressed in terms
of the difference of two squares, then it will be a saddle point because it will have
either positive or negative values depending on whether one of the numbers is larger
than the other.