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sesenic [268]
3 years ago
13

Write an exponential function y=ab^x whose graph passes through (1,6) and (3,54)

Mathematics
1 answer:
ehidna [41]3 years ago
7 0
y=ab^x\\\\(1;\ 6)\to x=1;\ y=6\\(3;\ 54)\to x=3;\ y=54\\\\\text{substitute the values of x and y to the equation}\\\\ \left\{\begin{array}{ccc}6=ab^1&\to&a=\dfrac{6}{b}^{(*)}\\54=ab^3\end{array}\right\\\\\text{substitute (*) to the second equation}\\\\54=\dfrac{6}{b}\cdot b^3\\\\54=6\cdot b^2\ \ \ \ |:6\\\\b=9\to b=\sqrt9\to b=3 \\\\\text{substitute the value of b to the (*)}\\\\a=\dfrac{6}{3}=2\\\\\text{Answer:}\ y=2\cdot3^x
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Determine consecutive integer values of x between which each real zero is located.
frozen [14]

Answer:

1. x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

2. x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

Step-by-step explanation:

Solve for x:

x^3 + 6 x^2 + 6 x - 4 = 0

The left hand side factors into a product with two terms:

(x + 2) (x^2 + 4 x - 2) = 0

Split into two equations:

x + 2 = 0 or x^2 + 4 x - 2 = 0

Subtract 2 from both sides:

x = -2 or x^2 + 4 x - 2 = 0

Add 2 to both sides:

x = -2 or x^2 + 4 x = 2

Add 4 to both sides:

x = -2 or x^2 + 4 x + 4 = 6

Write the left hand side as a square:

x = -2 or (x + 2)^2 = 6

Take the square root of both sides:

x = -2 or x + 2 = sqrt(6) or x + 2 = -sqrt(6)

Subtract 2 from both sides:

x = -2 or x = sqrt(6) - 2 or x + 2 = -sqrt(6)

Subtract 2 from both sides:

Answer: x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

_________________________________________

Solve for x:

x^4 - 2 x^3 - 6 x^2 + 8 x + 5 = 0

Eliminate the cubic term by substituting y = x - 1/2:

5 + 8 (y + 1/2) - 6 (y + 1/2)^2 - 2 (y + 1/2)^3 + (y + 1/2)^4 = 0

Expand out terms of the left hand side:

y^4 - (15 y^2)/2 + y + 117/16 = 0

Subtract -3/2 sqrt(13) y^2 - (15 y^2)/2 + y from both sides:

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (y^2 + (3 sqrt(13))/4)^2:

(y^2 + (3 sqrt(13))/4)^2 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

Add 2 (y^2 + (3 sqrt(13))/4) λ + λ^2 to both sides:

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (y^2 + (3 sqrt(13))/4 + λ)^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

-y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (2 λ + 15/2 + (3 sqrt(13))/2) y^2 - y + (3 sqrt(13) λ)/2 + λ^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = y^2 (2 λ + 15/2 + (3 sqrt(13))/2) - y + (3 sqrt(13) λ)/2 + λ^2

Complete the square on the right hand side:

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2 + (4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1)/(4 (2 λ + 15/2 + (3 sqrt(13))/2))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1 = 8 λ^3 + 18 sqrt(13) λ^2 + 30 λ^2 + 45 sqrt(13) λ + 117 λ - 1 = 0.

Thus the root λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2

Take the square root of both sides:

y^2 + (3 sqrt(13))/4 + λ = y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)) or y^2 + (3 sqrt(13))/4 + λ = -y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) + 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2))

Solve using the quadratic formula:

y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) + sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13))) - sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13))) or y = 1/4 (-sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) where λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3))

Substitute λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) and approximate:

y = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x - 1/2 = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x - 1/2 = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x - 1/2 = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or x = 1.67884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x = 1.67884 or x - 1/2 = 2.41497

Add 1/2 to both sides:

Answer: x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

8 0
3 years ago
Which of the following equations have exactly one solution?
Tems11 [23]

Answer:

A, B, & C

Step-by-step explanation:

A: -5x +12 = -12x - 12

    +12x        +12x

    7x + 12 = -12

         -12     -12

            7x = -24                 [Divide both sides by 7 to get x]

              x = -24/7

B: −5x + 12 = 5x + 12

   -5x            -5x

    -10x + 12 = 12

            -12     -12

            -10x = 0                   [Divide both sides by -10 to get x]

                 x = 0

C: −5x + 12 = 5x − 5

    -5x           -5x

    -10x + 12 = -5

            -12      -12

           -10x = -17                  [Divide both sides by -10 to get x]

               x = 17 / 10

D: −5x + 12 = −5x − 12

    +5x           +5x

              12 ≠ -12         [The statement is false, so it isn't the correct answer]

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3 years ago
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3 years ago
HELP PLEASE :)
Artist 52 [7]
Let b=boxes   SO 25=b(2.50)+3.98  So she would be able to buy 8.4 total boxes.
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3 years ago
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What is the domain, range and function of this, please help
GaryK [48]
Domain is your x values, Range is your y values.
4 0
3 years ago
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