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3 years ago

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If the original function F(x) = 2x^2 - 1 is shifted to the left 3 units to make the function g(x) , which expression would repre

sent g(x) ?
1 - 2(x-3)^2
2- 2( x+3) ^2 -1
3- 2x^2 + 2
4- 2x^2 - 4

1 answer:

aleksandrvk [35]3 years ago

4 0

That would be option 2
2(x + 3)^2 - 1

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Solution for: <br> -x + 3y -2z = 19<br> 2x + y - z = 5<br> -3x - y + 2z = -7

STatiana [176]

Answer:

x = -1 , y = 4 , z = -3

Step-by-step explanation:

Solve the following system:

{-x + 3 y - 2 z = 19 | (equation 1)

2 x + y - z = 5 | (equation 2)

-3 x - y + 2 z = -7 | (equation 3)

Swap equation 1 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

2 x + y - z = 5 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Add 2/3 × (equation 1) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y/3 + z/3 = 1/3 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Multiply equation 2 by 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Subtract 1/3 × (equation 1) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+(10 y)/3 - (8 z)/3 = 64/3 | (equation 3)

Multiply equation 3 by 3/2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+5 y - 4 z = 32 | (equation 3)

Swap equation 2 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+y + z = 1 | (equation 3)

Subtract 1/5 × (equation 2) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+(9 z)/5 = (-27)/5 | (equation 3)

Multiply equation 3 by 5/9:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add 4 × (equation 3) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y+0 z = 20 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 2 by 5:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add equation 2 to equation 1:

{-(3 x) + 0 y+2 z = -3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(3 x)+0 y+0 z = 3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 1 by -3:

{x+0 y+0 z = -1 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Collect results:

Answer: {x = -1 , y = 4 , z = -3

6 0

3 years ago