Find area of the isosceles triangle formed by the vertex and the x-intercepts of parabola y=x^2+4x−12.
+23245 To find the x-intercepts: y = x2 + 4x - 12
0 = x2 + 4x - 12
0 = (x + 6)(x - 2)
x-intercepts: x = -6 and x = 2
To find the vertex: y = x2 + 4x - 12
y + 12 = x2 + 4x
y + 12 + 4 = x2 + 4x + 4
y + 16 = (x + 2)2
vertex: (-2, -16)
The endpoints of the base of the isosceles triangle are (-6, 0) and (2, 0) ---> so its base is 8
The height of the triangle reaches from the midpoint of the base (-2, 0) and the vertex (-2, -16) ---> so its height is 16
The area is ½ · base · height = ½ · 8 · 16 = 64
