1. use the diagram to answer the following questions.
a) how many points appear in the figure?
b) how many lines appear in the figure?
c) how many planes appear in the figure?
d) name a line containing point v.
e) name the intersection of lines a and b.
f) give another name for line b.
g) name three non-collinear points.
h) give another name for plane d.

(a) points are; Y, X,V and Z

(b)a and b

(c) D

(d) a

(e) W

(f) line bisector.

(g) Y,X,W or VZW

(h)YXW or VZW

Step-by-step explanation:

(a)A point is an exact position on a plane surface usually indicated as a dot and has no size.In the figure, points are; Y, X,V and Z

(b) A line is a line of points which is extending infinitely in two directions.The lines appearing in this figure are; a and b

(c)A plane is a flat surface made of points which extend infinitely in all directions. The plane in the figure is D

(d)A line containing point V is that one which has the point V that extends infinitely in two directions. The line with point V is a

(e)An intersection point is where two lines meet at right-angle.The intersection point of line a and b is W

(f) A line that bisects another line segment into two equal parts is called a line bisector.

(g)Non-collinear points are those points which do not lie on the same line.In this case, points Y and X are non-collinear and points Z and V are also non-collinear

(h)A plane has two dimensions , extends without ending and its shape looks like that of a floor. Any three points not on the same line represents a plane.In this case, plane D is also called plane YXW

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 : )

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )