Question

PLEASE HELP!!!!! The expression 8x2 − 176x + 1,024 is used to approximate a small town's population in thousands from 1998 to 2018, where x represents the number of years since 1998. Choose the equivalent expression that is most useful for finding the year where the population was at a minimum.

8(x − 11)^2 − 56
8(x − 11)^2 + 56
8(x2 − 22x + 128)
8(x2 − 22x) + 128

Answer 1

Answer:

$8(x-11)^2+56$

Step-by-step explanation:

Given expression that represent the population from 1998 to 2018,

$8x^2-176x+1024$

Let,

$y=8x^2-176x+1024----(1)$

Which is a upward parabola,

Since, the minimum value of an upward parabola,

$y=a(x-h)^2+k$

is find at x = h,

From equation (1),

$y=8x^2-176x+1024$

$y=8x^2 - 176x + 968 - 968 + 1024$

$y=8(x^2-22x+121)+56$

$y=8(x-11)^2+56$

By comparing,

The population is minimum at x = 11. ( that is after 11 years since 1998 )

Hence, the equivalent expression that is most useful to find the year where population is minimum is,

$8(x-11)^2+56$

Answer 2

Answer:

$y=8(x-11)^2+56$

Step-by-step explanation:

We are given that an expression used to approximate small town's population in thousands from 1998 to 2018

$8x^2-176x+1024$

Where x represents the number of years

We have to find the equivalent expression which is most useful for finding the year where the population was at a minimum.

Let y=$8x^2-176x+1024$

$y=8(x^2-22x)+1024$

$y=8(x^2-2\times x\times 11+(11)^2-121)+1024$

$y=8(x^2-2\times x\times 11+(11)^2)-968+1024$

$y^2=8(x-11)^2+56$

By using identity

$(a-b)^2=a^2-2ab+b^2$

By compare with $y=a(x-h)^2+k$

(h,k)=Vertex of parabola

The value of y =k is minimum at x=h

We get (h,k)=(11,56)

Substitute x=11

$y=8(11-11)^2+56=56$

At x=11 the population  is minimum

Therefore,

$y=8(x-11)^2+56$

Hence, this is required expression which is equivalent to given expression and most useful for finding the year where the population was at a minimum.