Which best describes the range of the function f(x) = 2/3(6)x after it has been reflected over the x-axis?

2 answers:

7 0

Your post (" f(x) = 2/3(6)x ") would be clearer and less ambiguous if you'd please format it as follows:

f(x) = (2/3)(6)^x. The (2/3) shows that 2/3 is the coefficient of the exponential function 6^x. Please use " ^ " to indicate exponentiation.

Start by graphing f(x) = (2/3)(6)^x. The y-intercept, obtained by setting x=0, is (0, 2/3). Can you show that the value of f(x) is (2/3)*6, or 4, at x=1, (2/3)*6^2, or 24, at x = 2, and so on? What happens if x becomes increasingly smaller? The graph approaches, but does not touch, the x-axis.

If you complete this graphing assignment, then all you'd have to do is to flip the whole graph over vertically, reflecting it in the x-axis. You'll see that the graph never touchs the x-axis. Therefore, the range of this flipped graph is (-infinity, 0).

Leona [35]3 years ago

5 0

Answer:

The correct answer is:

D) all real numbers less than or equal to 0.

Step-by-step explanation:

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zlopas [31]

1. The function $h(t)=-16t^{2}+48t$ is a parabola of the form $ax^{2}+bx$ . The the formula for the axis of symmetry of a parabola is $x= \frac{-b}{2a}$ . We can infer from our function that $a=-16$ and $b=48$ , so lets replace those values in our formula:
$x= \frac{-b}{2a}$
$x= \frac{-48}{-2(16)}$
$x= \frac{-48}{-32}$
$x= \frac{3}{2}$
$x=1.5$
We can conclude that to the left of the line of symmetry the ball is reaching its maximum height, and to the right of the line of symmetry the ball is falling.

2. Lets check how much time the ball takes to reach its maximum height and return to the ground. To do that we are going to set the height equal to zero:
$0=-16t^{2}+48t$
$-16t(t-3)=0$
$t=0$ or $t-3=0$
$t=0$ or $t=3$
From our previous point we know that the ball reaches its maximum time at $t=1.5$ , which means that it takes 1.5 seconds to reach the maximum height and 1.5 seconds to fall back to the ground.