1.1 Review of functions  (Page 5/28)

1. To find the zeros, solve $f(x)=\mathrm{-4}x+2=0.$ We discover that $f$ has one zero at $x=1\text{/}2.$
2. The $y$ -intercept is given by $\left(0,f\left(0\right)\right)=\left(0,2\right).$
3. Given that $f$ is a linear function of the form $f\left(x\right)=mx+b$ that passes through the points $(1\text{/}2,0)$ and $(0,2),$ we can sketch the graph of $f$ ( [link] ).

   

1. To find the zeros, solve $\sqrt{x+3}+1=0.$ This equation implies $\sqrt{x+3}=\mathrm{-1}.$ Since $\sqrt{x+3}\ge 0$ for all $x,$ this equation has no solutions, and therefore $f$ has no zeros.
2. The $y$ -intercept is given by $\left(0,f\left(0\right)\right)=(0,\sqrt{3}+1).$
3. To graph this function, we make a table of values. Since we need $x+3\ge 0,$ we need to choose values of $x\ge \mathrm{-3}.$ We choose values that make the square-root function easy to evaluate.

   $\mathit{\text{x}}$	$\mathrm{-3}$	$\mathrm{-2}$	$1$
   $\mathit{\text{f}}\mathbf{(}\mathit{\text{x}}\mathbf{)}$	$1$	$2$	$3$

Making use of the table and knowing that, since the function is a square root, the graph of $f$ should be similar to the graph of $y=\sqrt{x},$ we sketch the graph ( [link] ).

1. 
   

   Height $s$ As a function of time $t$

   $\mathit{\text{t}}$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$
   $\mathit{\text{s}}\mathbf{(}\mathit{\text{t}}\mathbf{)}$	$100$	$96$	$84$	$64$	$36$	$0$

   Since the ball hits the ground when $t=2.5,$ the domain of this function is the interval $[0,2.5].$

2.