All categories

4 years ago

Graph the function f( x ) = 3√ x - 2+3

Mathematics

1 answer:

mezya [45]4 years ago

7 0

Step-by-step explanation:

Direction: Opens Up

Vertex:

(

−

)

(3,-3)

Focus:

(

−

)

(3,-114)

Axis of Symmetry:

x=3

Directrix:

−

y=-134

Select a few

x values, and plug them into the equation to find the corresponding

y values. The

x values should be selected around the vertex.

Tap for more steps...

−

xy112-23-34-251

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex:

(

−

)

(3,-3)

Focus:

(

−

)

(3,-114)

Axis of Symmetry:

x=3

Directrix:

−

y=-134

−

You might be interested in

What is a decrease of 39 point

Zolol [24]

I would think it would be -39 points

8 0

3 years ago

Write a number between 40 and 60. The number is divisible by 6 and has exactly 10 factors.

marshall27 [118]

There is 60 but only jas 9 facrors

6 0

3 years ago

Read 2 more answers

Anyone help? I'm trying to answer this question but I'm confused.

xeze [42]

N
Bbhhbjjbvfgbbyy Jill

3 0

3 years ago

Please help me solve this <br> 4(x-1)

JulijaS [17]

Simply multiply 4 for the terms in the parenthesis

4x-4

7 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago