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 :/
+9466 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)
+9466 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)
