### Video Transcript

Determine the variation function π£
of β for the function π of π₯ is equal to ππ₯ squared plus ππ₯ plus two at π₯ is
equal to one, and from π£ of one-half is equal to seven over two and π of one is
equal to six, determine the constants π and π.

In this question, weβre asked to
determine the variation function π£ of β for an unknown quadratic function π of π₯
at the value of π₯ is equal to one. This is the first part of the
question. So, letβs start by doing this. And to do this, weβre going to need
to recall what we mean by a variation function of a function π of π₯ at a specific
value. We can recall that the variation
function π£ of β for a function π of π₯ at π₯ is equal to π is given by π
evaluated at π plus β minus π evaluated at π. Itβs a measure of how much the
function changes when π₯ changes from π to π plus β.

And in our case, our function π of
π₯ is the quadratic ππ₯ squared plus ππ₯ plus two, and our value π is equal to
one. First, to determine the variation
function π£ of β, letβs start by substituting π is equal to one. We get that π£ of β is equal to π
evaluated at one plus β minus π evaluated at one. So, to find an expression for our
variation function, weβre going to need to substitute π₯ is equal to one plus β and
π₯ is equal to one into the given function π of π₯.

First, we substitute π₯ is equal to
one plus β into the function π of π₯. We get π multiplied by one plus β
squared plus π times one plus β plus two. We then need to subtract π
evaluated at one. So, we need to subtract π times
one squared plus π times one plus two. This gives us the following
expression for our variation function. And we now need to simplify this
expression. First, letβs start by distributing
the exponent over the parentheses in our first term. We can do this by using the
binomial formula or by using the FOIL method. In either case, we distribute this
to get one plus two β plus β squared, and we need to multiply this by π.

In our second term, we can
distribute the value of π over the parentheses. We multiply each term by π. π times one is π and π times β
is πβ. We then add two to this value, and
we can simplify further. We notice π times one squared is
equal to π and π times one is equal to π. Finally, we can distribute the
negative over these parentheses. So, we subtract π, we subtract π,
and we subtract two, giving us the following expression for π£ of β. And we can simplify this expression
further. We can notice that we have π minus
π, which is equal to zero, and we also have two minus two, which is also equal to
zero.

Letβs now distribute π over the
parentheses in our first term. We multiply each term inside the
parentheses by π. We get π plus two πβ plus πβ
squared, and we need to add πβ and subtract π. And we can continue
simplifying. We have π minus π, which is equal
to zero. And finally, we can notice the
first and third term in this expression share a factor of β. So, taking this factor out then
gives us that π£ of β is equal to πβ squared plus β multiplied by two π plus
π.

Letβs now clear some space and move
on to the second part of this question. We need to determine the values of
the constants π and π by using two pieces of information. π£ evaluated at one-half is equal
to seven over two and π evaluated at one is equal to six. So, letβs start with the first
equation weβre given. π£ evaluated at one-half is equal
to seven over two. And we can find an expression for
π£ evaluated at one-half by substituting β is equal to one-half into our equation
for π£ of β.

Substituting β is equal to one-half
into our function π£ of β gives us that π£ of one-half is equal to π times one-half
squared plus one-half multiplied by two π plus π. And now, we can simplify. First, one-half squared is equal to
one-quarter, so the first term is π over four. And in the second term, we can
distribute one-half over the parentheses to get π plus π over two. So, we have π over four plus π
plus π over two. And the first two terms π over
four plus π simplifies to give us five π over four. And remember, this needs to be
equal to seven over two.

So, we have one equation involving
π and π. Letβs now clear some space and find
the second equation by using the fact that π evaluated at one is equal to six. We have that six is equal to π
evaluated at one. And we can find an expression for
π evaluated at one by substituting π₯ is equal to one into π of π₯. This gives us that six is equal to
π times one squared plus π multiplied by one plus two. And we can evaluate this; it
simplifies to give us π plus π plus two. And remember, this needs to be
equal to six. However, we can subtract two from
both sides of the equation. And since six minus two is equal to
four, we get that four is equal to π plus π.

We now have a pair of simultaneous
equations in terms of π and π. And we can solve this in a few
different ways. Letβs start by clearing some
space. One way we can solve these
simultaneous equations is by noticing in the second equation, the coefficients of π
and π are both one. So, we can easily rearrange this
equation to find an expression for π in terms of π or π in terms of π. For example, if we subtract π from
both sides of the equation, we see that π must be equal to four minus π. We can then substitute π is equal
to four minus π into our first simultaneous equation. Doing this then gives us that seven
over two is equal to five times four minus π over four plus π over two.

We now have an equation entirely in
terms of π. So, we can solve for π by
isolating π on one side of the equation. To do this, letβs start by
distributing five over the parentheses. We multiply each term in the
parentheses by five. We get 20 minus five π, and we
need to divide all of this by four. So, we have seven over two is equal
to 20 minus five π all over four plus π over two. We can simplify our equation by
multiplying both sides of the equation by the lowest common multiple of the
denominators. In this case, thatβs four. So, we multiply both sides of the
equation by four to remove the denominators. Multiplying both sides of the
equation by four gives us that 14 is equal to 20 minus five π plus two π.

And now we can just solve this
equation for π. Negative five π plus two π is
negative three π. We can rearrange the equation to
get that three π is equal to six and then divide both sides of the equation through
by three. We see that π is equal to two. And we can verify this by
substituting π is equal to two back into our equation only in terms of π. We can also use this value of π to
determine the value of π since remember, π is equal to four minus π.

So, substituting π is equal to two
into this equation gives us that π is equal to four minus two, which simplifies to
give us two. And now that weβve shown that π is
equal to two and π is equal to two, we could substitute the values of these
constants into our function π£ of β. However, itβs not necessary. We can just say that π is two and
π is two.

Therefore, we were able to show the
variation function π£ of β for π of π₯ is equal to ππ₯ squared plus ππ₯ plus two
at π₯ is equal to one is π£ of β is equal to πβ squared plus β times two π plus
π. And if π£ of one-half is equal to
seven over two and π of one is equal to six, then π is equal to two and π is
equal to two.