### Video Transcript

Given that the domain of the function π of π₯ equals π over π₯ plus six over π₯ plus π is the real numbers minus the set containing the values negative four and zero and π of negative one equals two, find the values of π and π.

Weβve been given a rational function π of π₯, which is the sum of two algebraic fractions: π over π₯ and six over π₯ plus π. The values of π and π are currently unknown. And we need to determine them using the other information given in the question.

Letβs consider first how we can use the information about the domain of this function, which weβre told is the set of real numbers minus the values of negative four and zero. In general, the domain of a function is the set of all values of π₯ for which the function itself is well defined. We can think of it as the set of input values or the set of values on which the function acts.

Now, our function π of π₯ is a rational function. The only time we will be concerned about π of π₯ being undefined would be if we were trying to divide by zero, which would occur if either of the denominators were themselves equal to zero. So we can say that the function would be undefined if π₯ equals zero or if π₯ plus π is equal to zero. We can see that the value π₯ equals zero is excluded from the domain of the function. So this is consistent with our first equation.

The other value, which has been excluded from the domain, is negative four. So this must be the solution to our second equation. Substituting π₯ equals negative four into the second equation gives the equation negative four plus π is equal to zero. And adding four to each side, we find that π is equal to four. We see then that the function π of π₯ is in fact equal to π over π₯ plus six over π₯ plus four. And we can see why the values of zero and negative four are excluded from the domain of this function.

So we found the value of π, and now letβs consider how to find the value of π. To do this, weβll use the other piece of information we were given. π of negative one is equal to two. This means when we evaluate the function at negative one, so when we substitute π₯ equals negative one, the value we get is two. Substituting π₯ equals negative one gives π of negative one equals π over negative one plus six over negative one plus four. And we know that this is equal to two.

We can simplify this equation. π over negative one is equal to negative π. And in the denominator of the second fraction, negative one plus four is three. So we have negative π plus six over three is equal to two. Six over three is of course equal to two. So our equation has become negative π plus two equals two. We can then subtract two from each side of the equation, giving negative π is equal to zero. And if we multiply both sides of the equation by negative one, or just using a bit of logic, we find that π is equal to zero.

And so we found the values of both π and π. π is equal to four, and π is equal to zero.