Find the slope of the graph of the function \( f(x)=\sqrt{2 x} \) at \( (8,4) \). Then find an equation of the line tangent to the graph at that point. A. \( \lim _{h \rightarrow 0} \frac{\sqrt{2(8+h)}-\sqrt{2(8)}}{h} \) R. \( \lim _{h \rightarrow 0} \frac{\sqrt{2(8-h)}+\sqrt{2(8)}}{h} \) C. \( \lim _{h \rightarrow 0} \frac{\sqrt{8+h}-\sqrt{8}}{h} \) D. \( \lim _{h \rightarrow 0} \frac{\sqrt{2(8)}-\sqrt{2(8+h)}}{h} \) The slope of the graph of the function \( f(x)=\sqrt{2 x} \) at \( (8,4) \) is \( \square \) (Type an integer or a simplified fraction.)

We are given the function \[ f(x)=\sqrt{2x}. \] The slope of the tangent line at a point is given by the derivative \( f'(x) \). We will use the definition of the derivative at \( x=8 \) with the limit \[ \lim_{h \rightarrow 0} \frac{f(8+h)-f(8)}{h}. \] Substituting \( f(x) \) gives \[ \lim_{h \rightarrow 0} \frac{\sqrt{2(8+h)}-\sqrt{2(8)}}{h}. \] This matches option A. Now, we compute the derivative in general. Write the function in power form: \[ f(x)= (2x)^{\frac{1}{2}}. \] Differentiate using the chain rule: \[ f'(x)=\frac{1}{2}(2x)^{-\frac{1}{2}} \cdot 2. \] Simplify: \[ f'(x)= (2x)^{-\frac{1}{2}}=\frac{1}{\sqrt{2x}}. \] Now, evaluate the derivative at \( x=8 \): \[ f'(8)=\frac{1}{\sqrt{2(8)}}=\frac{1}{\sqrt{16}}=\frac{1}{4}. \] Thus, the slope of the graph at \( (8,4) \) is \[ \frac{1}{4}. \] Next, we find the equation of the tangent line at \( (8,4) \) using the point-slope form \[ y-y_1=m(x-x_1), \] where \( m=\frac{1}{4} \) and \( (x_1,y_1)=(8,4) \). This gives: \[ y-4=\frac{1}{4}(x-8). \] Solving for \( y \): \[ y=\frac{1}{4}x-2+4, \] \[ y=\frac{1}{4}x+2. \] In summary: - The correct limit definition is option A. - The slope of the tangent line is \( \frac{1}{4} \). - The equation of the tangent line is \( y=\frac{1}{4}x+2 \).