+0  
 
0
142
1
avatar

Graph A: y= x^2+4x

Graph B: y= 5

Graph C: y= 2x

 

a. Solve the equatation : x^2+4x= 0

b. What does this answer say about Graph A?

c. Calculate the coordinates of the points of intersection for A and B

d. Calculate the coordinates of the points of intersection for A and C

 

Also does anyone know how to actually calculate points of intersection? I always have difficulties with those kind of questions :/

Guest Mar 19, 2018

Best Answer 

 #1
avatar+7339 
+3

a.

x2 + 4x  =  0       Factor an  x  out of the terms on the left side.

x(x + 4)  =  0       Set each factor equal to zero.

 

x = 0     or     x + 4 = 0

                     x  =  -4

 

The solutions are   x = 0   and   x = -4  .

 

b.

It means that when  x = 0  and when  x = -4 ,  y = 0 .   The x-intercepts are  0  and  -4 .

 

c.

y  =  x2 + 4x

y  =  5

 

Since  y = 5  ,  we can plug in  5  for  y  in the first equation.

 

5  =  x2 + 4x               Subtract  5  from both sides.

0  =  x2 + 4x - 5          Factor the right side.

0  =  (x + 5)(x - 1)       Set each factor equal to zero.

 

x + 5  =  0        or        x - 1  =  0

x  =  -5             or        x  =  1

 

So the points of intersection for  A  and  B  are:   (-5, 5)   and   (1, 5)

 

d.

y  =  x2 + 4x

y  =  2x

 

Since  y = 2x , we can plug in  2x  for  y  in the first equation.

 

2x  =  x2 + 4x          Subtract  2x  from both sides.

0  =  x2 + 4x - 2x

0  =  x2 + 2x            Factor the right side.

0  =  x(x + 2)           Set each factor equal to zero.

 

x  =  0        or        x + 2 - 0

                             x  =  -2

 

When  x = 0 ,   y  =  2(0)  =  0     So one of the intersection points is   (0, 0)

When  x = -2 ,  y  =  2(-2)  =  -4     So the other intersection point is   (-2, -4)

 

The points of intersection for  A  and  C  are:  (0, 0)   and   (-2, -4)

hectictar  Mar 19, 2018
 #1
avatar+7339 
+3
Best Answer

a.

x2 + 4x  =  0       Factor an  x  out of the terms on the left side.

x(x + 4)  =  0       Set each factor equal to zero.

 

x = 0     or     x + 4 = 0

                     x  =  -4

 

The solutions are   x = 0   and   x = -4  .

 

b.

It means that when  x = 0  and when  x = -4 ,  y = 0 .   The x-intercepts are  0  and  -4 .

 

c.

y  =  x2 + 4x

y  =  5

 

Since  y = 5  ,  we can plug in  5  for  y  in the first equation.

 

5  =  x2 + 4x               Subtract  5  from both sides.

0  =  x2 + 4x - 5          Factor the right side.

0  =  (x + 5)(x - 1)       Set each factor equal to zero.

 

x + 5  =  0        or        x - 1  =  0

x  =  -5             or        x  =  1

 

So the points of intersection for  A  and  B  are:   (-5, 5)   and   (1, 5)

 

d.

y  =  x2 + 4x

y  =  2x

 

Since  y = 2x , we can plug in  2x  for  y  in the first equation.

 

2x  =  x2 + 4x          Subtract  2x  from both sides.

0  =  x2 + 4x - 2x

0  =  x2 + 2x            Factor the right side.

0  =  x(x + 2)           Set each factor equal to zero.

 

x  =  0        or        x + 2 - 0

                             x  =  -2

 

When  x = 0 ,   y  =  2(0)  =  0     So one of the intersection points is   (0, 0)

When  x = -2 ,  y  =  2(-2)  =  -4     So the other intersection point is   (-2, -4)

 

The points of intersection for  A  and  C  are:  (0, 0)   and   (-2, -4)

hectictar  Mar 19, 2018

23 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.