The equation g(x) in vertex form of a quadratic function for the transformations whose graph is a translation 4 units left and 1 unit up of the graph of f(x) is (x-4)² + 1

Given a quadratic function for the transformations given the function f(x) = x²

If the function g(x) of the graph is translated 4 units to the left, the equation becomes (x-4)² (note that we subtracted 4 from the x value

Translating the graph 1 unit up will give the final function g(x) as (x-4)² + 1 (We added 1 since it is an upward translation.)

Hence the equation g(x) in vertex form of a quadratic function for the transformations whose graph is a translation 4 units left and 1 unit up of the graph of f(x) is (x-4)² + 1

Learn more here: brainly.com/question/15381183

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2 years ago

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$