Mathematics, 13.05.2021 06:30 jeremiahharrisjr

Triangle DEF has vertices D = (-4,-4), E = (-2,-4), and F = (-3,-1). 5! 3 2 1 -7 6 5 4 3 2 2 3 4 5 6 10 5 3. What is the area of the triangle? Show your reasoning. 1. Plot the triangle in the coordinate plane and label the vertices. 2. Name the coordinates of 3 points that are inside the triangle. please help !!! <3

Triangle DEF has vertices D = (-4,-4), E = (-2,-4), and F = (-3,-1). 5! 3 2 1 -7 6 5 4 3 2 2 3 4 5

Answers: 1

Show answers

Answers

Answer from: kutiracing

Area = 3

Step-by-step explanation:

Given

D = (-4,-4)

E = (-2,-4)

F = (-3,-1)

Solving (a) and (b):

See attachment for plot and labelled vertices

Solving (c): The area

This is calculated using:

$Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2+x_3y_1 - x_1y_3|$

Where:

D = (-4,-4) -- (x_1,y_1)

E = (-2,-4) -- (x_2,y_2)

F = (-3,-1) -- (x_3,y_3)

This gives:

$Area = \frac{1}{2}|(-4*-4) - (-4*-2) + (-2 *-1) - (-3*-4)+(-3*-4) - (-4*-1)|$

$Area = \frac{1}{2}|16 - 8 +2- 12+12 - 4|$

$Area = \frac{1}{2}|6|$

This gives

$Area = \frac{1}{2} * 6$

Area = 3

Triangle DEF has vertices D = (-4,-4), E = (-2,-4), and F = (-3,-1). 5! 3 2 1 -7 6 5 4 3 2 2 3 4 5 6

Answer from: Quest

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.