Cho f(t)=(t-4)u(t-2)phep bien doi laplace la F(s)=e^-2s/s^2-2e^-2s/s62

What is the equation written in vertex from of a parabola with a vertex of (4, –2) that passes through (2, –14)?

vichka [17]

Answer:

y = -3(x - 4)² - 2

Step-by-step explanation:

Given the vertex, (4, -2), and the point (2, -14):

We can use the vertex form of the quadratic equation:

y = a(x - h)² + k

Where:

(h, k) = vertex

a = determines whether the graph opens up or down, and it also makes the parent function wider or narrower.

positive value of a = opens upward
negative value of a = opens downward
a is between 0 and 1, (0 < a < 1) the graph is wider than the parent function.
a > 1, the graph is narrower than the parent function.

h = determines how far left or right the parent function is translated.

h = positive, the function is translated h units to the right.
h = negative, the function is translated |h| units to the left.

k determines how far up or down the parent function is translated.

k = positive: translate k units up.
k = negative, translate k units down.

Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:

vertex (h, k): (4, -2)

point (x, y): (2, -14)

Substitute these values into the vertex form to solve for a:

y = a(x - h)² + k

-14 = a(2 - 4)² -2

-14 = a (-2)² -2

-14 = a4 + -2

Add to to both sides:

-14 + 2 = a4 + -2 + 2

-12 = 4a

Divide both sides by 4 to solve for a:

-12/4 = 4a/4

-3 = a

Therefore, the quadratic equation inI vertex form is:

y = -3(x - 4)² - 2

The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.

Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.

Please mark my answers as the Brainliest, if you find this helpful :)

8 0

2 years ago

Name two solutions to y = 3x - 5

wlad13 [49]

Answer:

(1,-2) and (0,-5) (There are an infinite amount more)

Step-by-step explanation:

The easiest way to find solutions is to plug in an x value. Let's try 1:

y = 3(1) - 5⇒y = 3 - 5⇒y = -2

(1,-2)

Let's try 0:

y = 3(0) - 5⇒y = 0 - 5 ⇒y = -5

(0,-5)

6 0

3 years ago

A store sells candles.

maria [59]

Answer:

25%

Step-by-step explanation:

If you start at 45% and it reduces by 20% you are just doing 45-20 which equals 25. (Please let me know if im wrong)

7 0

3 years ago

Consider the function f(x) = √5x + 15 for the domain [-3, ∞).Find f-¹(x), where f-1 is the inverse of f.Also state the domain of

Ymorist [56]

Answer: We have to find the inverse function of the following:

$f(x)=\sqrt{5x+15}\rightarrow[-3,\infty)\Rightarrow(1)$

The Inverse of the (1) is as follows:

$\begin{gathered} f(x)=y=\sqrt{5x+15} \\ \\ \text{ Switch: }x\text{ and }y\text{ and solve for }y(x)\text{ } \\ \\ \\ y=\sqrt{5x+15}\rightarrow x=\sqrt{5y+15} \\ \\ \\ x^2=5y+15 \\ \\ \\ x^2-15=5y \\ \\ \\ y=\frac{x^2}{5}-\frac{15}{5} \\ \\ \\ \\ y=f^{-1}(x)=\frac{x^2}{5}-3 \\ \\ \\ f^{-1}(x)=\frac{x^{2}}{5}-3 \end{gathered}$

The domain of the inverse function is:

$x\in(-\infty,+\infty)$

3 0

9 months ago

An AC generator produces an alternating current with a peak value of 18 V. The resistance of the lamp is 4 ohms. What is the eff

leva [86]

The effective current when an AC peak voltage of 18V is applied to a lamp with 4 ohm resistance is 3.18 A.

<h3>What is current?</h3>

Current is the flow of electrons through a conductor when voltage is applied across the conductor.

Given the peak voltage of 18 V, hence:

$V_{rms}=\frac{Peak\ value}{\sqrt{2} } =\frac{18}{\sqrt{2} } =12.73 V\\\\I_{rms}=\frac{V_{rms}}{resistance } =\frac{12.73}{4 } =3.18 A\\$

The effective current when an AC peak voltage of 18V is applied to a lamp with 4 ohm resistance is 3.18 A.

Find out more on current at: brainly.com/question/24858512

7 0

2 years ago