50 points me hurry

given the vertices of ∆abc are a (2,-5), b (-4,6) and c (3,1), find the vertices following each of the transformations from the original vertices:

rx = 3 t< 3,-6> r(90◦, o)

Q1)

Rx = 3

in this reflection, the image has been reflected along the x = 3 line. then all the points on the image are reflected along this vertical line. the distance between each point and the vertical line is noted and by that same distance its moved to the opposite side of the vertical line, while the y coordinates stay the same.

A(2,-5)- x coordinate is 2, its 1 unit away from x=3, it should move by 1 unit to the other side of x = 3, then x = 4. y stays as it is.

A' - (4,-5)

B(-4,6) - distance from x = 3, is (-4-+3) is seven units, moves by 7 units to the other side

B' (10,6)

C(3,1) = distance from 3 = 0, therefore the point stays at is,

C' (3,1)

Q2)

next is a translation, in translation the size nor shape changes, each point in the original image moves the same distance in the same direction.

T (3,-6) (x,y) = this means that the x coordinates move by +3 points to the right and y coordinates move downwards by 6 units

A (2,-5) ---> (2+3,-5-6)

A' (5,-11)

B (-4,6) ---> (-4+3,6-6)

B' (-1,0)

C (3,1) ---> (3 + 3, 1-6)

C' (6,-5)

Q3)

r(90°,0)

this is a rotation done in the clockwise direction. then the image takes a rotation to the right direction around the origin. 

then the x, y coordinates of the preimage become ;

(x,y) = (y,-x)

A(2,-5) --> A' = (-5,-2)

B(-4,6) ---> B' = (6,4)

C(3,1)  ---> C' = (1,-3)

Q1)

Rx = 3

in this reflection, the image has been reflected along the x = 3 line. then all the points on the image are reflected along this vertical line. the distance between each point and the vertical line is noted and by that same distance its moved to the opposite side of the vertical line, while the y coordinates stay the same.

A(2,-5)- x coordinate is 2, its 1 unit away from x=3, it should move by 1 unit to the other side of x = 3, then x = 4. y stays as it is.

A' - (4,-5)

B(-4,6) - distance from x = 3, is (-4-+3) is seven units, moves by 7 units to the other side

B' (10,6)

C(3,1) = distance from 3 = 0, therefore the point stays at is,

C' (3,1)

Q2)

next is a translation, in translation the size nor shape changes, each point in the original image moves the same distance in the same direction.

T (3,-6) (x,y) = this means that the x coordinates move by +3 points to the right and y coordinates move downwards by 6 units

A (2,-5) ---> (2+3,-5-6)

A' (5,-11)

B (-4,6) ---> (-4+3,6-6)

B' (-1,0)

C (3,1) ---> (3 + 3, 1-6)

C' (6,-5)

Q3)

r(90°,0)

this is a rotation done in the clockwise direction. then the image takes a rotation to the right direction around the origin. 

then the x, y coordinates of the preimage become ;

(x,y) = (y,-x)

A(2,-5) --> A' = (-5,-2)

B(-4,6) ---> B' = (6,4)

C(3,1)  ---> C' = (1,-3)

Rx=3

Step-by-step explanation:

Rx=3

Step-by-step explanation:

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )