### Video Transcript

The line π₯ minus π¦ minus three
equals zero touches the curve π¦ equals ππ₯ cubed plus ππ₯ squared at one,
negative two. Find π and π.

The key information given in this
question is that the line and the curve touch at this point with coordinates one,
negative two. But the line does not cross the
curve, which means that the line π₯ minus π¦ minus three equals zero is a tangent to
the given curve at this point. We know that the gradient of a
curve is equal to the gradient of the tangent to the curve at that point. The equation of our line is π₯
minus π¦ minus three equals zero. And when rearranging, we see that
this is equivalent to π¦ equals π₯ minus three. Comparing with π¦ equals ππ₯ plus
π, thatβs the general form of the equation of a straight line in slope-intercept
form, we see that the slope of our tangent is one. Can we find an expression for the
slope of the curve? Well, we can do this by
differentiation. By applying the power rule, we see
that dπ¦ by dπ₯ is equal to three ππ₯ squared plus two ππ₯.

Next, we evaluate this gradient
function at the point one, negative, two. So we substitute π₯ equals one into
our gradient function, giving three π plus two π. We can then equate the gradient of
the curve at this point with the gradient of the tangent to the curve at this
point. And it gives an equation involving
π and π: three π plus two π is equal to one. We canβt solve this equation
because we have only one equation and two unknowns. So weβre going to need to find a
second equation.

The point one, negative two lies on
both the curve and the tangent. So if we substitute the values of
one and negative two into the equation of the curve, weβll get a second equation
connecting π and π. We have π multiplied by one cubed
plus π multiplied by one squared is equal to negative two, simplifying to π and π
equals negative two. We now have two linear equations in
π and π, which we need to solve simultaneously. We can multiply equation two by
two, as this will make the coefficient of π the same as it is in equation one.

Weβll then subtract the second
equation from the first to eliminate the π terms, giving π equals five. Substituting this value for π into
our original equation two thatβs π and π equals negative two gives five plus π
equals negative two. And by subtracting five, we see
that π is equal to negative seven. So we found the values of π and
π. π is equal to five and π is equal
to negative seven.

A reminder is that the key point
that we used in this question is that the slope of a curve is equal to the slope of
the tangent to the curve at that point.