All categories

2 years ago

Given f(x)=x^2, write the equation of the transformed function if g(x)=f(x-1)

Mathematics

1 answer:

gayaneshka [121]2 years ago

4 0

Answer:

1. g(x) = (x - 1)²; translation of 1 unit to the right.

2. g(x) = 3^(x + 1) - 2; translation to the left of 1 unit and translation of 2 units down.

Step-by-step explanation:

f(x) = x²

g(x) = f(x - 1)

g(x) = (x - 1)²

Since x was replaced by x - 1, g(x) is similar to f(x) with a translation of 1 unit to the right.

f(x) = 3^x

g(x) = f(x + 1) - 2

g(x) = 3^(x + 1) - 2

Since x was replaced by x + 1, which is the same as x - (-1), there is a translation to the left of 1 unit. In addition, the -2 causes a translation of 2 units down.

You might be interested in

Nick had 3 1/4 bottles of water for his wrestling practice when he finished he had 124 bottles of water left he said he used to

Ostrovityanka [42]

Nick had 3 1/4 bottles before, and he had 1 2/4 bottles left, so that means
3 1/4 - 1 2/4 = 1.75
1.75= 1 3/4.
So, I do not agree with nick.

Hope this helps!!! :)

8 0

3 years ago

Write an equation in point-slope form of a line with a slope of –4 that goes through the point J(–5, 6).

Fittoniya [83]

The equation is y = -4x + 26.

5 0

3 years ago

What shape is produced when slicing a right rectangular pyramid perpendicular to the base

vovikov84 [41]

For the most part, the cross-section formed is a <em>trapezoid</em>, but if the slice passes through the apex of the pyramid, that shape is a <em>triangle</em>.

(Image source: MathCaptain.com)

5 0

3 years ago

Please I need help ASAP! I don't know any of the answers. For the second question, I accidentally clicked on it.

valkas [14]

PART A :
A,B,C,D
Because if the spinner was spinning on time it would have either landed on any of the given parts a,b,c,d once
PART B:
A and B
They are equally likely to occur because both of them have a portion of 90 degrees.
Hopefully my answer helped :)

3 0

2 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago