How does the graph of f(x)=-4^5x-3 differ from the graph of g(x)=-4^5x?

How does the graph of f(x)=-4^5x-3 differ from the graph of g(x)=-4^5x?


The answer should be 1. Reflection over the x-axis and shifted 3 units to the left.
Reflection over the y-axis and shifted 3 units up

Reflection over the x-axis and shifted 3 units to the left

Step-by-step explanation:

From the parent graph f(x) = -4^x, we can conclude that:

1. Reflected over the x-axis because we multiplied by -1 to flip it to positive.

2. Horizontally shifted the graph to the left because of the +3 (take the opposite for horizontal movement)

That is how we get f(x) = 4^{x+3}.

Edit: I have no idea why the mods and AL2006 reported my answer and deleted it. I still got the same answer. The child function f(x) = 4^{x+3}, not f(x) = 4^x+3

Option A is the answer

Step-by-step explanation:

Here f(x) =-4^x

in order to change the sign in front ,we need to  reflex it with respect to x axis

and since here 3 is added to x  [ as its x+3 in place of x ] which means  it change in x  which gives 3 units to the left.

  Option A is the answer.

A. f(x) = -4|x + 2| + 3

Step-by-step explanation:

An absolute value graph is a v-shaped graph whose equation has the form y = a| x - h| + k where (h,k) is the vertex. On the graph the vertex is (-2,3). This means its equation is y = a| x --2| + 3. It simplifies to y = a|x+2|+3. To find a, look at the answer options. Each option has 4 or -4. Since the graph faces downward, it has a negative leading coefficient of a = -4. The equation is y = -4|x+2|+3

\bf \boxed{y=a(x-{{ h}})^2+{{ k}}}\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
&\uparrow &\uparrow \\
\textit{now, you're asked for one with a vertex of -19, -18}\\
\textit{that simply means h = -19, and k= -18}
so, just change those h,k coordinates of the center :)

also, we're asked to put it in "standard form", that simply means, expand the squared binomial, and simplify, you'll end up with a trinomial

Do you know the answer?

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