Cases for a Quadratic Function

Cases for a quadratic function

There are three possibilities to consider, let's first eliminate the most complicated
one. Suppose 4ac – b2 is negative. If this is the case, then it means that what is
between the brackets is actually a positive quantity while the second term will be
negative times y2. So it will be a negative quantity. One term will be positive, and
one term will be negative. This tells us that we actually have a saddle point with our
function as a difference of two squares.

We could create new coordinates in which the value u equals x plus b over 2ay.
Using these coordinates, we can rewrite the equation as a difference of squares.
The second case is that if 4ac is equal to 0, then this term over here goes away, so
what we have is just a 1 by itself. What that means is actually that our function
depends only on one direction of things. In the other direction it's going to actually be
degenerate. So, for example, if I were to give you a function such as w = ×2 (where
w and x are two variables), it would mean that no matter what the value of y is, the
function will always equal ×2. If I were to graph this function on a coordinate plane, it
would look like a valley whose bottom is completely flat.
In other words, we have a degenerate critical point, meaning that there is a direction
along the y axis in which nothing happens.

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In fact. we have critical points everywhere
along the y axis. So whether the square of x is x or something else such as X plus
over 2ay (we don't really care), it will still get this degenerate behavior.
We will not try to determine precisely what happens in this case. We will accept that,
when we have a general function, rather than one that is quadratic, as in our
example, there will be other terms involving higher powers of x or y, and these will
obscure what happens in the valley.
[Graph]
In the case of a function of one variable, if you have just the function f(x) = ×5 and try
to decide what type of point (c,0) is, you will take the second derivative which is O
and then you cannot conclude because things depend on higher order derivatives.
So we won't like that case and won't try to figure out what's going on here. If,
however, 4ac – b2 is positive then that means that the big bracket up there equals a
sum of two squares which means that we have written was 1/4a times + something
squared plus something else squared.
According to the sign of a, this expression can take one of two forms. If a is positive,
the quantity (a + b)/2 will always be non-negative and so there exists a minimum. If a
is negative, the quantity will be negative if b is greater than or equal to -1 and will be
non-negative if -1 < b < 0; hence, there exists a maximum.
And if a is negative, on the other hand, that means that if we multiply this positive
quantity by a negative number, we get something that's always negative. So 0 is
actually the maximum.
The example shows that if w equals x squared, it does not depend on y. In other
words, a times x plus b over a times y squared is independent of x and y. This
means that if we choose to move in one direction-say, the perpendicular direction
where this remains constant- -then we stav at the minimum or maximum for some
values of x and y. So this direction is degenerate because nothing happens there.
We will now derive the second derivative of a given function. Start by assuming that
a, b, and c are known constants. The second derivative is what we're after and so
we do not need to compute derivatives yet. However, it should be noted that once a,
b and c are found as second derivatives of a function, we will have found their values
in terms of a and b.
A degenerate critical point is a point on a function at which the derivative is zero. If a
function has a degenerate critical point, it could be either a minimum or a maximum
depending on the sign of the second derivative. In general, when we have more
complicated functions, we cannot tell whether we have a degenerate critical point or
a saddle point.
A degenerate saddle point is possible to have, too. For example, if you have x cubed
plus y cubed, you can convince yourself that if you take x and y to be negative, it will
be negative. If x and y are positive, it's positive. And it has a very degenerate critical
point at the origin. So that's a degenerate saddle point. We don't see it here because
that doesn't happen if you have only quadratic terms like that. You need to have
higher order terms to see it happen.