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Artist 52 [7]
4 years ago
13

State the order and type of each transformation of the graph of the function ƒ(x) = 4(5x)3 – 6 as compared to the graph of the b

ase function
Mathematics
1 answer:
MaRussiya [10]4 years ago
3 0

Answer:

For the function, f(x) = 4(5x^3) -6, stretches the graph by making the outputs matched to the inputs much larger than the parent function f(x) = x^3. It stretches further by being multiplied by 4 after 5^3=125. Lastly the graph is shifted down by 6 units.

Step-by-step explanation:

When functions are transformed there are a few simple rules:

  • Adding/subtracting inside the parenthesis to the input shifts the function left(+) and right(-).
  • Adding/subtracting outside the parenthesis to the output shifts the function up(+) and down(-).
  • Multiplying the function by a number less than 1 compresses it towards the x-axis.
  • Multiplying the function by a number greater than 1 stretches it away from the x-axis.
  • Multiplying by a negative or changing the leading coefficient's sign will flip the graph.

For the function, f(x) = 4(5x^3) -6, stretches the graph by making the outputs matched to the inputs much larger than the parent function f(x) = x^3. It stretches further by being multiplied by 4 after 5^3=125. Lastly the graph is shifted down by 6 units.

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bixtya [17]

Answer:

the answer is a 0 b is x= -5 c is x = - 1.6 d is x= - 1.6

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Which is the equation of a parabola with a directrix at y=2 and a focus at (5,0)
CaHeK987 [17]

Check the picture below, so the parabola looks more or less like so.

the vertex is always half-way between the focus point and the directrix, and since the parabola is opening downwards, the "p" distance is negative.

\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=5\\ k=1\\ p=-1 \end{cases}\implies 4(-1)(y-1)~~ = ~~(x-5)^2\implies -4(y-1)=(x-5)^2 \\\\\\ y-1=-\cfrac{1}{4}(x-5)^2\implies y=-\cfrac{1}{4}(x-5)^2 + 1

4 0
3 years ago
ASAP TYWhich of the following is the perfect square of a binomial?
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Step-by-step explanation:

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Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

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Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

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Rzqust [24]

#Rate of change

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The x values of the x intercepts are the zeros

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