Answer to Question #155340 in Math for .

Question

a) given the parabola (y+2)²=-8(x + 1), sketch the graph and determine the vertex , focus and directrix line of the parabola .

b)sketch the graph of f(x)=-(|x - 1| - 3) by using transformation technique . begin with f(x).

c)graph the system inequality below and shade the feasible region .

: 3y < 15 - 2x

: y(less and equal than) x + 2

: y < 2

: y is (greater and equal than) 0

Answer 1

a) $(y+2)^2=-8(x+1)$

$h=-1, k=-2, p=-2$

$Vertex: (h, k)=(-1, -2)$

$Focus: (h+p, k)=(-3, -2)$

$Directrix: x=h-p, x=1$

b)

1. Begin by graphing $f_1(x)=|x|$ - black graph.

2. Shift left by 1: $f_2(x)=|x-1|$ - blue graph.

3. Shift down by 3: $f_3(x)=|x-1|-3$ - green graph.

4. Flip over the $x$ -axis: $f(x)=-(|x-1|-3)$ - red graph.

c)

$3y<15-2x=>y<-\dfrac{2}{3}x+5$

$y\leq x+2$

$y<2$

$y\geq0$

c)graph the system inequality below and shade the feasible region .

: 3y < 15 - 2x

: y(less and equal than) x + 2

: y < 2

: y is (greater and equal than) 0

$y=0: 0=15-2x=>x=\dfrac{15}{2}$

$Point (\dfrac{15}{2}, 0)$

$y=2: 3(2)=15-2x=>x=\dfrac{9}{2}$

$Point (\dfrac{9}{2}, 2)$

$y=0: 0=x+2=>x=-2$

$Point(-2,0)$

$y=2: 2=x+2=>x=0$

$Point(0,2)$