Question

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

Answer 1

To find the area of the isosceles triangle formed by the vertex and the x-intercepts of the parabola y=x^2+4x−12, we can graph the parabola, find the x-intercepts, and use Heron's formula to find the area of the triangle.

Step-by-step explanation:

To find the area of the isosceles triangle formed by the vertex and the x-intercepts of the parabola, we need to first graph the parabola and identify the x-intercepts. The equation of the parabola is y = x^2 + 4x - 12. To find the x-intercepts, we set y equal to zero and solve for x. So, 0 = x^2 + 4x - 12. We can factor this equation as (x - 2)(x + 6) = 0. Therefore, the x-intercepts are x = 2 and x = -6.

Now, we can construct the isosceles triangle using the vertex and the x-intercepts. The vertex of the parabola is given by the equation x = -b/2a. In this case, a = 1 and b = 4. So, x = -4/(2*1) = -2. The coordinates of the vertex are (-2, f(-2)). Now we have the three points (-2, f(-2)), (2, 0), and (-6, 0). We can use the distance formula to find the lengths of the triangle's sides.

Once we have the lengths of the triangle's sides, we can use Heron's formula to find the area. Heron's formula states that the area of a triangle with sides a, b, and c is given by the formula:

A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2. In this case, a = b (the length of the isosceles sides) and c = the base, which is the distance between the x-intercepts. Once you have the values for a, b, and c, substitute them into the formula to find the area of the triangle.

Answer 2

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

`(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
`(1)/(2)* \text { base }* \text { height }=(1)/(2)*8*16=64 \text{ sq units }`