The area of the isosceles triangle is 64 sq units.
<u>Solution:</u>
Part 1: x-intercepts
The x-intercepts occur at the points on the function where y=0
So, we need to solve
The left side factors fairly easily into:
So solution occur when
and
So the x-intercepts are at (0,6) and (0,−2)
Part 2: vertex of the parabola
The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to 0.
The derivative of the given quadratic is
By observation, this is equal to 0 when x=2
When x=2 the original equation becomes
Therefore the vertex of this parabola is at (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
