Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 57

$$y = \frac{1}{5}$$

Given $$f(x)=5^{x^{2}-2 x}, \quad x=1$$ Since at $x=1$, $y= 1/5$ and $$f'(x)= 5^{x^{2}-2 x} \ln 5 \frac{d}{dx} (x^{2}-2 x ) =\left(5^{x^{2}-2 x}\right)(\ln 5)(2 x-2)$$ Then $ m= f'(x)\bigg|_{x=1}=0$ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-1/5 }{x-1}&=0 \\ y&= \frac{1}{5} \end{align*}