2x^2-16x+15^i how do I find the axis of symmetry?

Question

anonymous · Answer

I don't know about with the imaginary exponent. but on a parabola you should be thinking about the vertex and its coordinates.

anonymous · Answer

well yeah it's the imaginary exponent that's confusing me

anonymous · Answer

http://www.askamathematician.com/2011/11/q-how-are-imaginary-exponents-defined/ im reading this trying to figure it out. I don't think its leading me anywhere though.

anonymous · Answer

@lonnie455rich thanks for the help!

anonymous · Answer

@Hero

ganeshie8 · Answer

its a typo may be ?

ganeshie8 · Answer

2x^2-16x+15 

see if u can find the axis-of-symmetry for this one.

ganeshie8 · Answer

based on ur previous question I am pretty sure ^i is a typo :o

anonymous · Answer

I was hoping it was a typo with my first answer.

anonymous · Answer

well without the I it's 4 but no it's not a typo

ganeshie8 · Answer

alright ! then complete the square

ganeshie8 · Answer

2x^2-16x+15^i

2(x^2-8x) + 15^i

2(x^2-2.4.x) + 15^i

2(x^2-2.4.x + 16 - 16) + 15^i

2(x^2-2.4.x + 16 ) + 15^i - 32

2(x-4)^2 + 15^i - 32
       ^

ganeshie8 · Answer

axis of symmetry is still x = 4

anonymous · Answer

:o you are a life saver!!! thank you soooo much!

ganeshie8 · Answer

i is just a number, lets treat it as such lol

ganeshie8 · Answer

np :)