Find area of the isosceles triangle formed by the vertex and the x-intercepts of parabola y=x2+4x−12.

   Area of the isosceles triangle is 64 square units.

   Equation of the parabola,

y = x² + 4x + 4 - 4 - 12

y = (x + 2)² - 4 - 12

y = (x + 2)² - 16

Vertex form of the parabola with vertex (h, k) is given by,

y = (x - h)² + k

By comparing both the equation, vertex of the parabola will be (-2, -16).

For x-intercepts of the parabola,

y = 0,

(x + 2)² - 16 = 0

(x + 2)² = 16

x + 2 = ±4

x = -6, 2

Therefore, vertices of the isosceles triangle are (-2, -16), (-6, 0) and (2, 0).

As shown in the graph,

Length of the base (AB) of the triangle = 2 - (-6) = 8 units

Height (CD) of the triangle = 16 units

Area of triangle =

                          =

                          =

                          = 64 square units

     Therefore, area of the triangle will be 64 square units.

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The area of the isosceles triangle is 64 sq units.

Solution:

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  

is there any attachment? what are the equations?