### Video Transcript

Find the equations to the tangent
lines of the curve π¦ equals π₯ plus eight multiplied by π₯ plus 10 at the points
where this curve intersects the π₯-axis.

Letβs begin, first of all, by
determining the coordinates of these points. Weβre looking for the points where
the curve intersects the π₯-axis. And we know that everywhere on the
π₯-axis, π¦ is equal to zero. So by setting our equation for π¦
equal to zero, we have an equation which we can solve to find the π₯-coordinates of
the points where the curve intersects the π₯-axis.

This is a quadratic equation
already in its factored form. So we can take each factor in turn,
set it equal to zero, and then solve the resulting linear equation, giving π₯ plus
eight equals zero, which leads to π₯ equals negative eight and π₯ plus 10 equals
zero, which leads to π₯ equals negative 10. The two points at which this curve
intersects the π₯-axis then are the points negative 10, zero and negative eight,
zero. We can sketch this curve, if we
want. Itβs a quadratic curve with a
positive leading coefficient intersecting the π₯-axis at negative 10 and negative
eight. So it looks a little something like
this.

Weβre then looking to find the
equations of the lines, which are tangent to this curve at the points where it
intersects the π₯-axis. So those are the two lines drawn in
green. We know that the general equation
of a straight line in its point-slope form is π¦ minus π¦ one equals π π₯ minus π₯
one, where π₯ one, π¦ one gives the coordinates of one point on the line and π
gives its slope. We know the coordinates of one
point on each of our tangent lines. So in order to apply this formula,
we need to determine their slopes. We can do this by recalling that
the slope of the line tangent to a curve at any given point is the same as the slope
of the curve itself at that point, which we can find by evaluating its first
derivative.

Before we get into differentiating,
letβs manipulate our expression for this curve slightly by distributing the
parentheses and then simplifying to give π¦ is equal to π₯ squared plus 18π₯ plus
80, a polynomial. We recall then that the power rule
of differentiation, which tells us that the derivative with respect to π₯ of a
general power term ππ₯ to the πth power for real values of π and π is equal to
ππ multiplied by π₯ to the power of π minus one. We multiply by the exponent and
then reduce the exponent by one.

We can now apply this power rule of
differentiation to find the derivative of our function π¦. The derivative of π₯ squared with
respect to π₯ is two π₯. The derivative of positive 18π₯ is
positive 18, which we can see if we think of 18π₯ as 18π₯ to the power of one. And finally, the derivative of a
constant term positive 80 is just zero, which we can see if we think of 80 as 80π₯
to the power of zero. We find then that the general slope
function of our curve dπ¦ by dπ₯ is equal to two π₯ plus 18.

We then need to evaluate the slope
at each of the points weβre interested in. First of all, when π₯ is equal to
negative 10, the slope will be equal to two multiplied by negative 10 plus 18. Thatβs negative 20 plus 18 which is
equal to negative two. Secondly, when π₯ is equal to
negative eight, the slope will be equal to two multiplied by negative eight plus
18. Thatβs negative 16 plus 18 which is
equal to two. Notice that this is consistent with
what weβve seen on our sketch. The tangent line at the point where
π₯ equals negative 10 has a negative slope. Whereas the tangent to the point
when π₯ equals negative eight has positive slope.

Next, we can use the general
equation of our straight line. At the point with coordinates
negative 10, zero where the slope is equal to negative two, we have the equation π¦
minus zero equals negative two multiplied by π₯ minus negative 10. That simplifies to π¦ equals
negative two multiplied by π₯ plus 10. And distributing the parentheses,
we have π¦ equals negative two π₯ minus 20. We can then collect all the terms
on the left-hand side of the equation to give the equation of this tangent as π¦
plus two π₯ plus 20 is equal to zero.

For our other tangent at the point
negative eight, zero where the slope is equal to two, we have the equation π¦ minus
zero equals two multiplied by π₯ minus negative eight. That simplifies to π¦ equals two
multiplied by π₯ plus eight. And then distributing the
parentheses gives π¦ equals two π₯ plus 16. Again, collecting all terms on the
left-hand side of the equation gives the equation of this tangent line in the form
π¦ minus two π₯ minus 16 is equal to zero.

So weβve completed the problem and
found the equations of both lines, which are tangent to this curve at the points
where it intersects the π₯-axis. They are π¦ plus two π₯ plus 20
equals zero and π¦ minus two π₯ minus 16 equals zero.