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3 years ago

15

f(x) = √x. and g(x)=8√x Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical str

etch.

1 answer:

Snowcat [4.5K]3 years ago

7 0

Answer:

vertical stretch

Step-by-step explanation:

the 8 is outside the sq. root function, meaning it causes a vert. stretch

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30 POINTS!!!!!!!! JoeNah, DeKobian and Evan wrote equations to calculate the amount of money in a savings account after one year

castortr0y [4]

Answer:

See the explanation for the answer to this question

Step-by-step explanation:

Let

B ---> the balance of money saved.

M ---> the amount invested in dollars

JoeNah's Equation

M=$100

$B=1.5(100)=\$150$

Dekobian's Equation

M=$100

$B=100+0.05(100)=\$105$

Evan's Equation

M=$100

$B=M(1+0.05)$

Substitute

$B=100(1+0.05)=\$105$

Dekobian's equation and Evan's equation are correct! :)

JoeNah's equation is wrong. :(

Hope this helps you!

Have a good evening!

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3 years ago

What type of equation is The area of a square is a function of the length of the side of the square.

Lana71 [14]

Answer:

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Step-by-step explanation:

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3 years ago

WILL GIVE BRAINLIEST!

wel

D, E and F will lie on the line

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3 years ago

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What is x+9=2(x_1)^2 in the form of ax^2+bx +c=0

Naddik [55]

<span>x + 9 = 2(x - 1)^2
x + 9 = 2(x^2 - 2x + 1)
x + 9 = 2x^2 - 4x + 2
</span>2x^2 - 4x + 2 - x - 9 = 0
2x^2 - 5x -7 =0

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3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers