SIERENCE

A student says that the graph of the equation y = 3(x + 8) is the same as the graph of y = 3x, only translated upwards by 8 unit

Wittaler [7]

Given:

A student says that the graph of the equation $y = 3(x + 8)$ is the same as the graph of $y = 3x$ , only translated upwards by 8 units.

To find:

Whether the student is correct or not.

Solution:

Initial equation is

$y=3x$

$f(x)=3x$

Equation of after transformation is

$y=3(x+8)$

$g(x)=3(x+8)$

Now,

$g(x)=f(x+8)$ ...(i)

The translation is defined as

$g(x)=f(x+a)+b$ ...(ii)

Where, a is horizontal shift and b is vertical shift.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

From (i) and (ii), we get

$a=8\text{ and }b=0$

Therefore, the graph of $y=3x$ translated left by 8 units. Hence, the student is wrong.

8 0

2 years ago

I need help urgent plz someone help me solved this problem! Can someone plz help I’m giving you 10 points! I need help plz help

VMariaS [17]

Answer: a) 8.779 years

b) 8.664 years

<u>Step-by-step explanation:</u>

a)

$A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}$

A: accumulated amount (balance)
P: principal amount (original/initial investment)
r: interest rate (convert to a decimal)
n: number of times compounded per year
t: number of years

Given: A = 1800, P = 900, r = 8% = 0.08, n = 3, t = unknown

$1800=900\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\2=\bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=ln \bigg(1+\dfrac{0.08}{3}\bigg)^{3t}\\\\\\ln\ 2=3t\ ln\bigg(1+\dfrac{0.08}{3}\bigg)\\\\\\\dfrac{ln\ 2}{3\ ln\bigg(1+\dfrac{0.08}{3}\bigg)}=t\\\\\\\large\boxed{8.779=t}$