#### Answer

y=4x-3

#### Work Step by Step

The question asks us to find the equation of the tangent of the function
$f(x)=x^2 + 1$ at the point $(2,5)$
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:
$f'(x)=2x$
$f'(2)=4$
Thus, the gradient of that function at the value x=2 is 4.
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 $(2,5)$ and $(0,c)$ (where c is the y-intercept).
$\frac{5-c}{2-0}=4$
Solving for c gives us $c=-3$
Thus, the equation of the tangent at $(2,5)$ is $y=4x-3$