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NemiM [27]
2 years ago
14

1.consider the function

Mathematics
1 answer:
marin [14]2 years ago
4 0

Function transformation involves changing the form of a function

  • The transformation from f(x) to g(x) is a horizontal shift 3 units left, followed by a vertical stretch by a factor of 4
  • The x and y intercepts are -0.67 and 1.91, respectively.
  • The behavior of g(x) is that, g(x) approaches infinity, as x approaches infinity.

The functions are given as:

\mathbf{f(x) = log3x}

\mathbf{g(x) = 4log3(x+1)}

<u>(a) The transformation from f(x) to g(x)</u>

First, f(x) is shifted left by 1 unit.

The rule of this transformation is:

\mathbf{(x,y) \to (x + 1,y)}

So, we have:

\mathbf{f'(x) = log3(x + 1)}

Next. f'(x) is vertically stretched by a factor of 4.

The rule of this transformation is:

\mathbf{(x,y) \to (x,4y)}

So, we have:

\mathbf{g(x) = 4log3(x+1)}

Hence, the transformation from f(x) to g(x) is a horizontal shift 3 units left, followed by a vertical stretch by a factor of 4

<u>(b) Sketch of g(x)</u>

See attachment

<u>(c) Asymptotes</u>

The graphs of g(x) have no asymptote

<u>(d) The intercepts, and the behavior of f(x)</u>

The graph crosses the x-axis at x =-0.67, and it crosses the y-axis at y = 1.91

Hence, the x and y intercepts are -0.67 and 1.91, respectively.

The behavior of g(x) is that, g(x) approaches infinity, as x approaches infinity.

We know this because, the value of the function increases as x increases

Read more about function transformations at:

brainly.com/question/13810353

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A cone has a volume of 5 cubic inches. What is the volume of a cylinder that the cone fits exactly inside of?
DENIUS [597]
The answer is 15 in³.

The volume of the cone is:
V_1= \pi r_1 ^{2}  \frac{h_1}{3} = \frac{ \pi  r_1^{2} h_1}{3}
where:
V₁ - the volume of the cone
r₁ - the radius of the cone
h₁ - the height of the cone

The volume of the cylinder is:
V_2= \pi  r_2h_2^{2}
where:
V₂ - the volume of the cone
r₂ - the radius of the cone
h₂ - the height of the cone

Since <span>the cone fits exactly inside of the cylinder, they have the same radius and the height:
r</span>₁ = r₂
h₁ = h₂

Also:
V₁ = 5

Now, let's write two volume formulas together:
V_1= \frac{ \pi r^{2} h}{3}
<span>V_2= \pi rh^{2}</span>

We can include V₂ into V₁:
V_1= \frac{V_2}{3}

⇒ V_2=3*V_1
V_2=3*5 in^{3}
V_2=15 in^{3}
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