### Video Transcript

Determine the equation of the line
tangent to the curve π¦ equals four π₯ cubed minus two π₯ squared plus four at the
point negative one, negative two.

So weβve been given the equation of
a curve. And we need to determine the
equation of the line that is tangent to this curve at a particular point. Weβre going to use the formula for
the general equation of a straight line π¦ minus π¦ one equals π π₯ minus π₯
one. We already know the coordinates π₯
one, π¦ one. Itβs the point negative one,
negative two. But what about π, the slope of
this line? Well, we recall that the slope of a
curve is equal to the slope of the tangent to the curve at that point. So in order to find the slope of
this tangent, weβre first going to find the gradient function of the curve dπ¦ by
dπ₯.

We can do this by applying the
power rule of differentiation, giving dπ¦ by dπ₯ equals four multiplied by three π₯
squared minus two multiplied by two π₯. Remember, a constant differentiates
to zero. So that plus four just
differentiates to zero in our derivative, which simplifies to 12π₯ squared minus
four π₯. Now, thatβs the general gradient
function of this curve. But we want to know the gradient at
a particular point. So we need to evaluate dπ¦ by dπ₯
when π₯ is equal to negative one because thatβs our π₯-coordinate at this point. This gives 12 multiplied by
negative one squared minus four multiplied by negative one, which simplifies to
16.

We now know that the slope of this
tangent is 16 and the coordinates of a point that it passes through are negative
one, negative two. So we have all the information we
need in order to use the formula for the general equation of a straight line. Substituting the values of π, π₯
one, and π¦ one gives π¦ minus negative two equals 16 π₯ minus negative one. Thatβs π¦ plus two equals 16π₯ plus
16. And, then, subtracting two from
each side in order to collect the constants gives π¦ equals 16π₯ plus 14. So this is the equation of the line
tangent to the given curve at the point negative one, negative two. It passes through the point
negative one, negative two and it has the same gradient as the curve at that
point.