Imaginary and quadratics

Find the vertex for

 $f\left(x\right)\ =\ 2\left(x-6\right)^2-18$  

• (-6,-18)

• (6,-18)

• (-6,18)

• (6,18)

What are the x-intercepts for  $f\left(x\right)\ =\ 2\left(x-6\right)^2-18$  

• -9 and -3

• -9 and 3

• 9 and 3

• 9 and -3

What is the y-intercept for y =2(x-6)²-18 ?

• 54

• -30

• 90

• -54

Simplify (6-7i) - (-2 + 3i)

• 8-4i

• 8-10i

• 4-4i

• 9+32i

Simplify (6-7i)(-2+3i)

• 9+32i

• -33+32i

• 9-4i

• -33-32i

 $\frac{8}{5+2i}$  

Simplify

•  $\frac{40}{21}-\frac{16}{21}i$  

•  $\frac{40}{21}+\frac{16}{21}i$  

•  $\frac{40}{29}-\frac{16}{29}i$  

•  $\frac{40}{29}+\frac{16}{29}i$  

Simplify (6-7i)²

• -13

• 85

• 85-84i

• -13-84i

 $6\sqrt{-3}-3\sqrt{-12}$  

Simplify

•  $12i\sqrt{3}$  

•  $4i\sqrt{3}$  

•  $2i\sqrt{3}$  

• 0

 Solve the following equation and put your answer in a+bi form.   $x^2+4x+7=0$ 

•  $-2\pm2i\sqrt{3}$  

•  $-2\pm i\sqrt{3}$  

•  $-4\pm i\sqrt{3}$  

•  $-4\pm2i\sqrt{3}$  

An archer's arrow follows a parabolic path. The height of the arrow can be modeled by f(x) = -.005x²+2x+5, where f(x) is in feet. What is the maximum height of the arrow?

• 200 ft

• .005 ft

• 5.01 ft

• 205 feet