The slope of the line tangent to f(x) at x=5 can be found through:
[tex]m_{\tan }=\lim _{h\to0}\frac{f(x_0+h)-f(x_0)}{h}[/tex]
this is the same definition as the derivative of the function
then,
[tex]\begin{gathered} f^{\prime}(x)=2\cdot-8x \\ f^{\prime}(x)=-16x \end{gathered}[/tex]
then, evaluate at the given point
[tex]\begin{gathered} f^{\prime}(5)=-16\cdot5 \\ f^{\prime}(5)=-80 \end{gathered}[/tex]
the slope of the line tangent to f(x) at x=5 is -80.
Also, we know that the instantaneous rate of change is the slope of the tangent line, meaning the instantaneous rate of change is -80 as well.
The equation of the tangent line is given by
[tex]y-y_0=m\cdot(x-x_0)[/tex]
then, find f(5)
[tex]\begin{gathered} f(5)=-8\cdot(5)^2+8 \\ f(5)=-200+8 \\ f(5)=-192 \end{gathered}[/tex]
the equation of the line is
[tex]\begin{gathered} y-(-192)=-80(x-5) \\ y+192=-80x+400 \\ y=-80x+400-192 \\ y=-80x+208 \end{gathered}[/tex]
