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# Graph questions

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Graph A: y= x^2+4x

Graph B: y= 5

Graph C: y= 2x

a. Solve the equatation : x^2+4x= 0

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 :/

Mar 19, 2018

#1
+7543
+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)

Mar 19, 2018

#1
+7543
+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