Question

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

Answer 1

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

$x^2-4x-12=0$

The left side factors fairly easily into:

$(x-6)(x+2)=0$

So solution occur when

$x-6=0\rightarrow x=6$

and

$x+2=0\rightarrow x=(-2)$

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

$\frac{dy}{dx}=2x-4$

By observation, this is equal to 0 when x=2

When x=2 the original equation becomes

$y=(2)^2-4(2)-12$

$y=-16$

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)

$\Rightarrow$ so its base is 8

The height of the triangle reaches from the midpoint of the base (-2, 0) and the vertex (2, -16)

$\Rightarrow$ so its height is 16

The area is  $\frac{1}{2}\times \text { base }\times \text { height }=\frac{1}{2}\times8\times16=64 \text{ sq units }$