Which set of points represents a function? { ( 1 , − 2 ) , ( 3 , − 5 ) , ( 3 , − 2 ) , ( 5 , − 7 ) , ( 7 , − 8 ) } { ( 1 , − 2 ) , ( 3 , − 5 ) , ( 3 , − 2 ) , ( 5 , − 7 ) , ( 7 , − 8 ) } { ( 0 , 1 ) , ( 1 , − 3 ) , ( 2 , − 5 ) , ( 3 , − 7 ) , ( 4 , − 9 ) } { ( 0 , 1 ) , ( 1 , − 3 ) , ( 2 , − 5 ) , ( 3 , − 7 ) , ( 4 , − 9 ) } { ( − 1 , 1 ) , ( − 1 , 3 ) , ( − 1 , 5 ) , ( − 1 , 7 ) , ( − 1 , 9 ) } { ( − 1 , 1 ) , ( − 1 , 3 ) , ( − 1 , 5 ) , ( − 1 , 7 ) , ( − 1 , 9 ) } { ( − 4 , 0 ) , ( − 3 , − 1 ) , ( − 3 , 1 ) , ( 0 , − 2 ) , ( 0 , 2 ) } { ( − 4 , 0 ) , ( − 3 , − 1 ) , ( − 3 , 1 ) , ( 0 , − 2 ) , ( 0 , 2 ) }

Given:

The set of points.

To find:

Which set of points represents a function?

Solution:

A set of points represents a function if there exist a unique y-value for each x-value.

It option A, the set of points is

{(1,-2), (3,-5), (3,-2), (5,-7), (7, -8)}

This is not a function because there exist two y-values y=-5 and y=-2 for x=3.

It option B, the set of points is

{(0,1),(1,−3),(2,−5),(3,−7),(4,−9)}

This is a function because there exist a unique y-value for each x-value.

It option C, the set of points is

{(−1,1),(−1,3),(−1,5),(−1,7),(−1,9)}

This is not a function because there exist five y-values y=1,3,5,7,9 for x=-1.

It option D, the set of points is

{(−4,0),(−3,−1),(−3,1),(0,−2),(0,2)}

This is not a function because there exist two y-values y=-1 and y=1 for x=-3.

Therefore, the correct option is B.

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 : )

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2