### Video Transcript

If the variation function π of π₯
is equal to ππ₯ squared plus ππ₯ at π₯ is equal to π is given by π of β is equal
to πβ squared plus πβ, what is the value of π?

In this question, weβre given a
quadratic function π of π₯ which is equal to ππ₯ squared plus ππ₯. And weβre told when π₯ is equal to
π, the variation function of this quadratic function π of π₯ is given by π of β
is equal to πβ squared plus πβ for some unknown constants π and π. We need to use this information to
determine the value of π.

To do this, letβs start by
recalling what we mean by the variation function of a function π of π₯ at π₯ is
equal to π. We can recall itβs the function π
of β which is equal to π evaluated at π plus β minus π evaluated at π. In other words, this function
measures how much the function π changes when its input values change from π to π
plus β. And in this question, weβre given
an expression for the variation function π of β. Itβs πβ squared plus πβ.

We can substitute this expression
into the left-hand side of the equation. And we can also find an expression
for the right-hand side of this equation by noting weβre given the function π of
π₯. Itβs ππ₯ squared plus ππ₯. So we can substitute π₯ is equal to
π plus β and π₯ is equal to π into this function to find an expression for the
right-hand side of this equation. First, if we substitute π₯ is equal
to π plus β into the function π of π₯, we get π times π plus β all squared plus
π multiplied by π plus β. Then, we need to subtract π
evaluated at π, which we can find by substituting π₯ is equal to π into π of
π₯. We subtract ππ squared plus
ππ. This then gives us an equation in
terms of π, β, π, and π.

Letβs now rearrange this
equation. Letβs start by distributing the
exponent over the parentheses. And we can do this by using the
binomial formula or by expanding the brackets π plus β multiplied by π plus β. We get π squared plus two πβ plus
β squared. And we need to multiply this
expression by π. We can also distribute the negative
over the two terms at the end of this expression. This gives us that πβ squared plus
πβ is equal to π times π squared plus two πβ plus β squared plus π times π
plus β minus ππ squared minus ππ.

We can simplify the right-hand side
further by distributing the factor of π and the factor of π over the
parentheses. Doing this gives us the following
equation, which we can simplify. We have ππ squared minus ππ
squared, which is equal to zero. And we also have π times π minus
π times π, which is equal to zero. We can also note that we have πβ
squared on both the left- and right-hand side of the equation and π times β on both
the left- and right-hand side of the equation. So we can subtract πβ squared from
both sides of the equation and π plus β from both sides of the equation. This gives us that zero is equal to
two times π times π multiplied by β.

So, for this equation to hold true,
one of our unknowns must be equal to zero. Itβs worth noting this canβt be β
because β is a variable. It can take any value for our
variation function. And π is a parameter of our
original equation; itβs not equal to zero. Otherwise, our original function
would not be quadratic. So the only way for this equation
to hold true for all values of β would be for π to be equal to zero. Therefore, we were able to show if
the variation function of π of π₯ is equal to ππ₯ squared plus ππ₯ at π₯ is equal
to π is π of β is equal to πβ squared plus πβ, then the value of π must be
zero.