Pls help

1 answer:

8 0

The difference between g(x) and f(x) is that g(x) has a 3 in front being multiplied.

g(x) is the same as saying y, so…
y=3(x+2)2

Because it is multiplying all of the terms by 3, y will be 3 times larger than it would be in f(x). Y is the vertical variable, so it will be stretched vertically by 3.

Alternatively, depending on how your teacher teaches it, 3 is in the location of the “a” value. A is a number that stretches an equation vertically by a factor of itself. Since a=3, the equation is stretched by a factor of 3.

You might be interested in

The equation for line p can be written as y= – 5/4 x–6. Line q includes the point (5, – 8) and is parallel to line p. What is th

Elden [556K]

Answer:

y = -(5/4)x - (1 3/4)

Step-by-step explanation:

A line parallel to y= -(5/4)x - 6 will have the same slope, -(5/4), but a different value for the y-intercept, which is -6 in this equation. This means we can write:

y = -(5/4)x + b

As klong a b is not -6, the line will be parallel. But we are told that point (5,-8) must be on the new line. That means we need to select a value of b to make that happen. By shifting b, we can move the straight line up or down. To move it so that it goes through (5,8), enter that point in the equation and solve for b:

y = -(5/4)x + b

-8 = -(5/4)*5 + b for (5,-8)

-8 = -(25/4) + b

b = -8 + (25/4)

b = -(32/4)+(25/4)

b = -(7/4) or - 1 3/4

The parallel line that goes through (5,-8) is y = -(5/4)x - 2

5 0

1 year ago

For which rational expression is –2 an excluded value of x?

Kipish [7]

Answer:

option a is the answer

Step-by-step explanation:

put the value of x in each expression as x = -2

you'll see that in option a the expression becomes

$\frac{x - 3}{ {x}^{2} - 4} = \frac{ - 2 - 3}{ {( - 2)}^{2} - 4} \\ = \frac{ - 5}{4 - 4} \\ = \frac{ - 5}{0}$

which is not defined. hence -2 is excluded value

5 0

2 years ago

What is -2 1/3- 1 1/3

romanna [79]

Answer:

Step-by-step explanation:

$-2\dfrac{1}{3}-1\dfrac{1}{3}=-2\dfrac{1}{3}+\left(-1\dfrac{1}{3}\right)=-\left(2\dfrac{1}{3}+1\dfrac{1}{3}\right)\\\\=-\bigg[(2+1)+\left(\dfrac{1}{3}+\dfrac{1}{3}\right)\bigg]=-\left(3+\dfrac{1+1}{3}\right)\\\\=-3\dfrac{2}{3}$