If G(x)=1/x were shifted 3 units to the left and 3 units up, what would the new equation be?
2 answers:
Answer:
Option C will be the new equation.
Step-by-step explanation:
If a function is the form of f(x) = ax
and we know when a function is shifted left by b then variable x becomes [x - (-b)] and if the this function is shifted upwards by c then it gets the form of
f(x) = a(x + b) + c
Similarly in the given question parent function is shifted left by 3 and up by 3 units.
And the parent function is
Now the new function will become as
Therefore Option C will be the correct option.
You might be interested in
The first step for solving this expression is to insert what a and b stand for into the expression. This will change the expression to the following:
2(2)(4)³ + 6(2)³ - 4(2)(4)²
Now we can start solving this by factoring the expression
2(2 × 4³ + 3 × 2³ - 4 × 4²)
Write 4³ in exponential form with a base of 2.
2(2 ×
+ 3 × 2³ - 4 × 4²)
Calculate the product of -4 × 4².
2(2 × + 3 × 2³ -4³)
Now write 4³ in exponential form with a base of 2.
2(2 × + 3 × 2³ -)
Collect the like terms with a base of 2.
2( + 3 × 2³)
Evaluate the power of 2³.
2( + 3 × 8)
Evaluate the power of.
2(64 + 3 × 8)
Multiply the numbers.
2(64 + 24)
Add the numbers in the parenthesis.
2 × 88
Multiply the numbers together to find your final answer.
176
This means that the correct answer to your question is option A.
Let me know if you have any further questions.
:)
2(2)(4)³ + 6(2)³ - 4(2)(4)²
Now we can start solving this by factoring the expression
2(2 × 4³ + 3 × 2³ - 4 × 4²)
Write 4³ in exponential form with a base of 2.
2(2 ×
Calculate the product of -4 × 4².
2(2 ×
Now write 4³ in exponential form with a base of 2.
2(2 ×
Collect the like terms with a base of 2.
2(
Evaluate the power of 2³.
2(
Evaluate the power of
2(64 + 3 × 8)
Multiply the numbers.
2(64 + 24)
Add the numbers in the parenthesis.
2 × 88
Multiply the numbers together to find your final answer.
176
This means that the correct answer to your question is option A.
Let me know if you have any further questions.
:)