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kkurt [141]
3 years ago
11

I am a bit confused on this one, not really sure where to go with it other than the fact that you change T(x) to zero

Mathematics
1 answer:
bekas [8.4K]3 years ago
7 0
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)
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Answer:

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<u>8</u>x <u>- 12</u>y = <u>-24</u>

| | |

A B C

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8x - 12y = -24

-8x -8x

(First through the subtraction property of equality, remove 8x from both sides so that -12y is by itself on the left)

-12y = -8x - 24

×-1 ×-1 ×-1

(Through the identity property of negative 1, remove the negative sign from all of the numbers because a negative times a negative is a positive)

12y = 8x + 24

(Lastly, through the division property of equality, divide all sides by 12 because it is the coefficient of y, which will solve for the variable)

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2 years ago
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The appropriate choice is ...
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3 years ago
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