Write an equation for each translation of y=|x|. 2.5 units left y = | x | – 2.5

Mathematics

2 answers:

ch4aika [34]2 years ago

8 0

For this case we have that the main function is given by:

$y = f (x) = | x |$

We want to move the function 2.5 units to the left.

For this, we follow the following rule:

$g (x) = f (x + 2.5)$

Therefore, according to the rule we have:

$g (x) = | x + 2.5 |$

Answer:

The equation of the translated function is given by:

$g (x) = | x + 2.5 |$

Option C

Zigmanuir [339]2 years ago

6 0

<span>for every function like f(x), f(x+a) brings the diagram of a units left
y = | x + 2.5 | </span>

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Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

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So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

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So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

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$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$