The diagram shows part of the curve $y = x^{\tfrac{3}{2}}$ and $y = -x^{\tfrac{3}{2}}$. The tangent meets the curve $y = x^{\tfrac{3}{2}}$ at the point A where $x = 4$.

1. Find the coordinates of point A on the curve.

At A, $x=4$ and $y=x^{\tfrac32}$.

$y=4^{\tfrac32}=\left(\sqrt{4}\right)^3=2^3=8$

So $A=(4,8)$.

2. Find the gradient of the tangent at A.

For $y=x^{\tfrac32}$,

$\frac{dy}{dx}=\frac32 x^{\tfrac12}$

At $x=4$,

$\frac{dy}{dx}=\frac32(4^{\tfrac12})=\frac32(2)=3$

So the gradient of the tangent is $3$.

3. Form the equation of the tangent line through A.

Using $y-y_1=m(x-x_1)$ with $m=3$ and $(x_1,y_1)=(4,8)$,

$y-8=3(x-4)$

$y-8=3x-12$

$y=3x-4$

1. Substitute $x=1$ into the equation of the curve containing B.

Since B lies on $y=-x^{\tfrac{3}{2}}$ and its x-coordinate is $1$,
$y=-(1)^{\tfrac{3}{2}}=-1.$