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3 years ago

Which rule should be applied to reflect f(x) = x^3 over the x axis?

Mathematics

2 answers:

rjkz [21]3 years ago

6 0

Answer:

B. Multiply f(x) by -1

Step-by-step explanation:

When we want to reflect over the x axis f(x) becomes -f(x)

so x^3 becomes -x^3

Brut [27]3 years ago

4 0

Answer:

B. Multiply f(x) by -1.

Step-by-step explanation:

By multiplying f(x) by -1. That is:

$g(x) = - x^{3}$

Hence, the right choice is B.

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What is the value of x? <br> a. 2/3b) 4/3c) 4d) 6?

oksian1 [2.3K]

Value of x=6....
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8 0

3 years ago

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

solong [7]

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

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

switch both sides

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

subtract 3 from both sides

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

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{1}{2}=\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$ ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

$\sqrt{-1}=i$

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

$x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

4 0

2 years ago

Answer 2 questions about equations A and B.

Otrada [13]

Answer:

Step-by-step explanation:

I'm not sure how to answer the 2. question

8 0

3 years ago

Find the solution to the inequality 2x - 6 ≥ 3x -4 _________.

Alona [7]

Answer:x<=-2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Drag each factor to the correct location on the image. If p(1) = 3, p(-4) = 8, p(5) = 0, p(7) = 9, p(-10) = 1, and p(-12) = 0, d

Dominik [7]

Answer:

Step-by-step explanation:

Recall that when we are factoring a polynomial, we are implicitly finding its roots. By finding a root, we are looking for a value of x (say c) such that p(c) =0. When this happens, our polynomial has the the polynomial (x-c) as a factor.

We are given the values of p at some values of x. We notice that p(5) = 0 and p(-12) =0. So, this means that our polynomial has as a factor the polynomials (x-5) and (x+12).

4 0

3 years ago