Which is one of the transformations applied to the graph of f(x) = x2 to change it into the graph of g(x) = –3x2 – 36x – 60?

Mathematics

1 answer:

FromTheMoon [43]3 years ago

8 0

Hey There!! ☄️ The general vertex to The general vertex form is this:

v(x) = a (x-h)2 + k

where (h,k) is the coordinates of the of vertex.

and a indicates the widening or shrinking of the function compared to another parabolic function. If a become bigger, the graph becomes narrower. If a becomes negative, the graph is reflected over the x-axis.

Comparing f(x) = x2 with g(x) = -3(x+6)2 + 48, we have the following transformations:

The graph is reflected over the x-axis

The graph is made narrower.

The graph is shifted 6 units to the left.

The graph is shifted 48 units up.

From the choices we only have:

The graph of f(x) = x2 is made narrower

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Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)

dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = $\int_a^b \delta\left|r^{\prime}(t)\right| d t$

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$