How does the graph of g(x) = (x − 1)3 + 5 compare to the parent function f(x) = x3?

Mathematics

2 answers:

9966 [12]3 years ago

7 0

Answer:

the third one is your answer

Fynjy0 [20]3 years ago

3 0

Answer:

Step-by-step explanation:

Given:

- The original function is:

f(x) = x^3

Find:

How does the graph of g(x) = (x − 1)3 + 5 compare to the original function.

Solution:

- We have a general form of the new function relating to parent function.

g(x) = a*( x +/- b )^n + c

Where; a, b and c are constants.

- The constant a magnitude denotes steepness of the graph relative to 1 and the sign of a will determine the mirror image of the graph about line y = 0.

- The constant b magnitude denotes shifts of the graph of every x value sign of b will determine the direction of shifts. + b : shift left , - b shift right.

- The constant c magnitude denotes shifts of the graph of every y value. sign of c will determine the direction of shifts. + c : shift up , - b shift down.

- In our g(x). a = 1 , b = -1 , c = + 5

Hence, g(x) is shifted 1 unit to the right and 5 units up.

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The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is

Ray Of Light [21]

Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.

An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.

Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.

Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.

Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.

To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.

For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false

For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true

For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true

For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false

For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false

For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.

4 0

3 years ago

Please help!! Find BD

Temka [501]

Answer: $8\sqrt{2}$

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Work Shown:

Focus entirely on triangle ABD (or on triangle BCD; both are identical)

The two legs of this triangle are AB = 8 and AD = 8. The hypotenuse is unknown, so we'll say BD = x.

Apply the pythagorean theorem.

$a^2 + b^2 = c^2\\\\c = \sqrt{a^2 + b^2}\\\\x = \sqrt{8^2 + 8^2}\\\\x = \sqrt{2*8^2}\\\\x = \sqrt{8^2*2}\\\\x = \sqrt{8^2}*\sqrt{2}\\\\x = 8\sqrt{2}\\\\$

So that's why the diagonal BD is exactly $8\sqrt{2}\\\\$ units long

Side note: $8\sqrt{2} \approx 11.3137$

6 0

3 years ago

Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. after c years, calvin has $658.80. makayl

vodomira [7]

You will need a special formula to compute this.
Years = log (total/principal) / [n * log * (1 + rate / n)]

Part A) Calvin $400 5% monthly 658.80 Time = ?
Monthly compounding "n" = 12
Years = log(658.80/400) / [12 * log(1+ (.05/n))
Years = log ( 1.647 ) / (12 * log ( 1.0041666667 )
Years = 0.21669359917 / 12 * 0.0018058008777
Years = 0.21669359917 / 0.0216696105 
Years = 9.999884362 


Part B) Makayla 300 6% quarterly 613.04Time=?
Quarterly compounding
n = 4
Years = log (total/principal) / [n * log * (1 + rate / n)]
Years = log (613.04/300) / [4 * log (1 + .06/4)]
Years = log ( 2.0434666667 ) / 4 * log (1.015)
Years = 0.31036755784 / 4 * 0.0064660422492
Years = 0.31036755784 / 0.025864169 
Years = 11.9999044949 

So, the difference is roughly 3 years.