Which function represents a reflection of f(x) = 5(0.8)x across the x-axis?

Mathematics

1 answer:

lukranit [14]2 years ago

8 0

The function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

<h3>Reflection of functions and coordinates</h3>

Images that are reflected are mirror images of each other. When a point is reflected across the line y = x, the x-coordinates and y-coordinates change their position. In a similar manner, when a point is reflected across the line y = -x, the coordinates <u>changes position but are negated.</u>

Given the exponential function below

f(x) = 5(0.8)^x

If the function f(x) is reflected over the x-axis, the resulting function will be

-f(x)

This means that we are going to negate the function f(x) as shown;

f(x) = -5(0.8)^x

Hence the function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

Learn more on reflection here: brainly.com/question/1908648

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The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

$y=0.00035x^{2}$

Compute the first order derivative of <em>y</em> as follows:

$y=0.00035x^{2}$

$\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]$

$=2\cdot 0.00035x\\\\=0.0007x$

Now, it is provided that |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 605

Compute the arc length as follows:

$\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx$

$=\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\$

Now, let

$x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u$

$\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u$

$={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}$

Plug in the solved integrals in Arc Length and solve as follows:

$\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\$