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WINSTONCH [101]

2 years ago

6

A quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (−5,−3). What is the vertex of the

function defined as g(x)=−f(x)+3?

1 answer:

Licemer1 [7]2 years ago

4 0

Answer:

(-5 , 6)

Step-by-step explanation:

f(x) = (x - (-5))² - 3 ... vertex (h , k) : (-5 , -3) y = (x-h)² + k

g(x) = -f(x) + 3 = -((x+5)² - 3) + 3 = -(x + 5)² + 6

vertex of g(x) is : (-5 , 6)

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Evaluate the surface integral. s (x + y + z) ds, s is the parallelogram with parametric equations x = u + v, y = u − v, z = 1 +

Stolb23 [73]

$S$ is given to be parameterized by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u+v,u-v,1+2u+v\rangle$

with $0\le u\le3$ and $0\le v\le2$ . We have

$\mathbf r_u=\langle1,1,2\rangle$
$\mathbf r_v=\langle1,-1,1\rangle$
$\mathbf r_u\times\mathbf r_v=\langle3,1,-2\rangle$
$\left\|\mathbf r_u\times\mathbf r_v\right\|=\sqrt{14}$

The surface integral is then

$\displaystyle\iint_S(x+y+z)\,\mathrm dS=\iint_S(x(u,v)+y(u,v)+z(u,v))\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}((u+v)+(u-v)+(1+2u+v))\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}(4u+v+1)\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
$=48\sqrt{14}$

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