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LUCKY_DIMON [66]
3 years ago
6

The function f(x)=(x-1)^4 is not one to one. However, if we restrict the domain to x greater than or equal to 0, we can find it'

s inverse. Which Graph is the graph of f^-1(x) ?

Mathematics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

Option A

Step-by-step explanation:

Domain should be greater than/equal to 1 for the function to be one-one.

f has vertex at (1,0)

f inverse has (0,1)

f(x) = 4

(x - 1)⁴ = 4

x = 2.414

On f: (2.414, 4)

On f inverse: (4, 2.414)

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Help me solve this.<br>​
almond37 [142]

Answer:

thx-step explanation:

6 0
3 years ago
What is the equation in vertex form of the quadratic function with a vertex at (-1, -4) that goes through (1, 8)?
cestrela7 [59]

Answer:

y = 3(x+1)^2 - 4

Step-by-step explanation:The general form of the equation of a quadratic function whose vertex is (h,k) and whose leading coefficient is a is:

y - k = a(x-h)^2, or

y      = a(x-h)^2 - k

Substituting the coefficients of the vertex (-1, -4), we get:

y      = a(x + 1)^2 - 4

Substituting the coordinates of the given point, (1,8), we get:

8      = a(1+1)^2 - 4, which simplifies to:

8      = a(2)^2 - 4, or

8  = 4a - 4.  Then 4a = 12, and a = 3.

Thus, the desired equation is y = 3(x+1)^2 - 4 (answer j).


5 0
3 years ago
What is the value of y?​
prohojiy [21]

Answer:

it is 108 because 180-72

8 0
2 years ago
Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer
Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

0=(3x+2)(x-4)

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply (3x+2)(x-4) i.e. 3x^2-12x+2x-8 = 3x^2-10x-8

So, the equation becomes:

0=3x^2-10x-8

Comparing this equation to standard quadratic equation:

ax^2+bx+c=0

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}

The answer is same from both the approaches.

So, the correct answer is:

C) Both

3 0
2 years ago
Please please help me!!
Alona [7]

Answer:

g(x) = f(x+1) + 1

Step-by-step explanation:

Since the graph is only moved you don't have to worry about the slope.

To move the graph to the left you add to the x and in this case, it only moved 1 over.

to move the graph up you add to the whole function and in this case, it only moved up 1.

4 0
3 years ago
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