Semester 1 Practice Final

For the graph shown, the domain is

• $\left(-\infty,\ +\infty\right)$

• $x>0$

• $y\ge1$

• $\left(0,+\infty\right)$

The interval  $4\le x<10$  can be written in interval notation as

•  $\left[4,\ 10\right]$  

•  $\left[4,\ 10\right)$  

•  $\left(4,\ 10\right)$  

•  $\left(4,\ 10\right]$  

How is the graph of $g\left(x\right)=x^2-5\$ transformed as compared to  $f\left(x\right)=x^2$ ?

•  $g\left(x\right)$  is moved 

  
  

  right 5 units

•  $g\left(x\right)$  is moved 

  
  

  down 5 units

•  $g\left(x\right)$  is moved 

  
  

  left 5 units

•  $g\left(x\right)$  is moved 

  
  

  up 5 units

Which equation shows a transformation to $f\left(x\right)=x^3$ of 2 units right, 4 units down, and a reflection over the x-axis?

•  $g\left(x\right)=\left(x+2\right)^3+4$  

•  $g\left(x\right)=-\left(x+2\right)^3-4$  

•  $g\left(x\right)=-\left(x-2\right)^3-4$  

•  $g\left(x\right)=-\left(x-2\right)^3+4$  

Which table is the inverse of the function in the given table

• 
• 
• 
• 

What is the inverse of the function
 $f\left(x\right)=2x+4$   ?

•  $f^{-1}\left(x\right)=4\left(x-2\right)$  

•  $f^{-1}\left(x\right)=\frac{x-4}{2}$  

•  $f^{-1}\left(x\right)=\frac{x}{2}-4$  

•  $f^{-1}\left(x\right)=\frac{x}{4}+2$  

What is the vertex of the graph of

 $g\left(x\right)=-2\left|x+3\right|-4$  

•  $\left(-4,\ -3\right)$  

•  $\left(-2,\ -4\right)$  

•  $\left(3,\ -4\right)$  

•  $\left(-3,\ -4\right)$  

Consider the graph of: $g\left(x\right)=5\left|x-6\right|+4$  Which is NOT a transformation as compared to the graph of  $f\left(x\right)=\left|x\right|$  ?

• reflected over the x-axis

• translated 6 units to the right

• translated 4 units up

• vertically stretched by a factor of 5

solve the quadratic equation.

 $2x^{2\ }-7=81$  

•  $\pm\sqrt{2}$  
  

•  $\pm\sqrt{88}$  
  

•  $\pm2\sqrt{11}$  
  

•  $\pm11\sqrt{2}$  

Solve the quadratic equation.

 $5x^2-2=-6$  

•  $\pm\frac{2}{5}i$  

•  $\pm\frac{\sqrt{2}}{\sqrt{5}}i$  

•  $\pm\frac{2}{\sqrt{5}}i$  

•  $\pm\frac{2}{\sqrt{5}}$