What your teacher wanted you to do was to find where the equation equaled zero I have no idea why she would want yu do to that though because the turning point on the graph would be the vertex, not the zeroes of the graph
using math we do math stuff this equation is 3rd degree T(x)=(x+5)^3+7 I hope you meant (x+5)^2+7, because then the answer would be at (-5,7)
if yo meant 3rd degree then here wer go with calculus and deritivives
take the derivieve and find where the slope is 0 T'(x)=3(x+5)^2 where it is equal 0 is wehre x=-5 use sign chart it is positive at both places no turning point ?????? well, dunno
if it was 2nd degree euation then vertex or turning point would be at (-5,7)
First, you need to isolate w. So you get w = 2/3 divided by 4. I made the 2/3 a decimal, as it is easier to work with: 0.6666 divided by 4 is 0.167. 0.167 is 1/6. w = 1/6, or 0.167.
