The graph of g(x) is f(x) translated to the left 8 units and up 2 units. What is the function rule for g(x) given f(x) = x²?

1 answer:

7 0

Translation of a function y = h(x) to the right/left is given by y = h(x + a) for translation to the left by 'a' units and y = h(x - a) for translation to the right by 'a' units

Translation of a function y = h(x) up/down the y-axis is given by y = h(x) + a for translation 'a' unit up and y = h(x) - a for translation 'a' unit down

So, translating f(x) = x² eight units right and two units up gives:

g(x) = (x-8)² + 2

You might be interested in

Malkia is thinking of a mystery number. She cuts it in half then adds 5. The result is 24. Determine the mystery number. I need

White raven [17]

Answer:38

Step-by-step explanation:if you take 24 and subtract the five she added, it gives you 19 and since she divided it by two you would multiply it by two to get the original number she was thinking of

Hope this helps :)

6 0

3 years ago

Read 2 more answers

What is the sixth term of the sequence shown in the table

horsena [70]

Answer:

375

Step-by-step explanation:

trust me

4 0

3 years ago

Read 2 more answers

What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .