Answer
$y=12x-16$
Work Step by Step
The question asks us to find the equation of the tangent of the function
$h(t)=t^3$ at the point (2,8)
Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function.
Using the power rule:
$h(t)=t^3$
$f′(x)=3t^{3-1}=3t^2$
$f′(2)=3(2)^2=12$
Thus, the gradient of that function at the value x=2 is 12.
Using this information, we can find the intercept value of the tangent line.
The equation of the gradient is given by:
$\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$
We can now use the points (2,8) and (0,c) (where c is the y-intercept).
$\frac{8−c}{2-0}=12$
Solving for c gives us c=-16
Thus, the equation of the tangent at (2,8) is $y=12x-16$
