### Video Transcript

For the vector-valued function π« of π‘ is equal to π‘ squared plus 4, seven π‘ minus five, four π‘ cubed minus two π‘, evaluate π« at negative three.

The question gives us a vector-valued function, π« of π‘, and we can see that this function outputs a column vector. We need to find π« evaluated at negative three. This is not the usual form which we see vector-valued function given in. Normally, theyβre given in terms of the unit vectors π’, π£, and π€. So, one way of answering this question could be to rewrite our function in terms of the vectors π’, π£, and π€. In general, if we have a vector-valued function π« of π‘ which is equal to the column vector π₯ of π‘, π¦ of π‘, π§ of π‘, then π« of π‘ is also equal to π₯ of π‘ times π’ plus π¦ of π‘ times π£ plus π§ of π‘ times π€. Where π’, π£, and π€ are our standard unit directional vectors. We represent the fact that theyβre unit vectors with a hat.

And in fact, thereβs a lot of different notation used for vectors. For example, we use the half-arrow notation and the hat notation in this video. However, you might also see full-arrow notation, bold notation, or underlined letters. These all mean roughly the same thing. So, letβs apply this to the vector-valued function π« of π‘ given to us in the question. We have π₯ of π‘ is π‘ squared plus four, π¦ of π‘ is seven π‘ minus five, and π§ of π‘ is four π‘ cubed minus two π‘. So, in our case, π« of π‘ is equal to π‘ squared plus four π’ plus seven π‘ minus five π£ plus four π‘ cubed minus two π‘ π€.

The question is asking us to find the value of π« evaluated at negative three. So we just need to substitute π‘ is equal to negative three into this expression. Substituting π‘ is equal to negative three into this expression, we get π« of negative three is equal to negative three squared plus four π’ plus seven times negative three minus five π£ plus four times negative three cubed minus two multiplied by negative three π€. Now, all we need to do is evaluate each of the coefficients of our unit directional vectors. And if we do this, we get 13π’ minus 26π£ minus 102π€.

Normally, we would leave our answer like this. However, remember, originally π« of π‘ was given as a column vector, so we should rewrite our answer as a column vector. To do this, remember the coefficient of π’ will be the first entry of our column vector, the coefficient of π£ will be the second entry in our column vector, and the coefficient of π€ will be the third entry in our column vector. So, this gives us that π« of negative three is equal to the column vector 13, negative 26, negative 102.

As an aside, itβs worth noting we couldβve just worked with our column vectors directly. We just need to substitute π‘ is equal to negative three directly into our column vector. Doing this, we get π« of negative three is equal to the column vector negative three squared plus four, seven times negative three minus five, four times negative three cubed minus two multiplied by negative three. And now, all we have to do is evaluate each component of our column vector separately. And if we do this, we get the column vector 13, negative 26, negative 102. And this is exactly the same answer we got using our other method.

Therefore, weβve seen two different methods to evaluate the vector-valued function π« of π‘ is equal to the column vector π‘ squared plus four, seven π‘ minus five, four π‘ cubed minus two π‘ when π‘ is equal to negative three. Both of these answers gave us the column vector 13, negative 26, negative 102.