Without graphing, describe the shape of the graph of the function. Find the second coordinates of the points with first coordinates 0 and 1 . \( f(x)=e^{-1.8 x} \) The graph Find the second coordinates of the given first coordinates. \[ \begin{array}{rl}x & f(x)=e^{-1.8 x} \\ 0 & f(0)=\square \\ 1 & f(1)=\square\end{array} \] (Round to one decimal place as needed.)

Suppose that the functions \( f \) and \( g \) are defined as follows. \[ \begin{array}{l}f(x)=2 x+3 \\ g(x)=\sqrt{4 x+1} \\ \text { Find } \frac{f}{g} \text { and } f-g \text {. Then, give their domains using interval notation. } \\ \begin{array}{l}\left(\frac{f}{g}\right)(x)=\square \\ \text { Domain of } \frac{f}{g}: \square \\ \text { Domain of } f-g: \square\end{array} \\ \qquad\end{array} \] \( \begin{array}{l}\qquad-g)=\square\end{array} \) \[ \]

Consider the following. \[ \begin{array}{l}P(x)=\frac{1-5 x}{1-x} \\ \text { (a) State the domain of the function. } \\ \text { all real numbers } x \text { except } x=\frac{1}{5} \\ \text { all real numbers } x \text { except } x=-\frac{1}{5} \\ \text { all real numbers } x \text { except } x=-1 \\ \text { all real numbers } x\end{array} \] all real numbers \( x \) except \( x=1 \)

257. Funksiýalaryň grafigini guruń: \( \begin{array}{ll}\text { a) } y=x^{2}+6 x ; & \text { c) } y=0,5 x^{2}-4 \\ \text { b) } y=-\frac{1}{2} x^{2}+2 ; & \text { d) } y=\frac{1}{2} x^{2}-3 x\end{array} \)

257. Funksiýalaryň grafigini guruñ: \( \begin{array}{ll}\text { a) } y=x^{2}+6 x ; & \text { c) } y=0,5 x^{2}-4 \\ \text { b) } y=-\frac{1}{2} x^{2}+2 ; & \text { d) } y=\frac{1}{2} x^{2}-3 x\end{array} \)

Unit 3 AsSessment Path of Lava Lava coming from the eruption of a volcano follows a parabolic path. The height \( h \) in feet of a piece of lava \( t \) seconds after it is ejected from the volcano is given by \( h(t)=-t^{2}+16 t+936 \).