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
[tex]x^2-4x-12=0[/tex]
The left side factors fairly easily into:
[tex](x-6)(x+2)=0[/tex]
So solution occur when
[tex]x-6=0\rightarrow x=6[/tex]
and
[tex]x+2=0\rightarrow x=(-2)[/tex]
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
[tex]\frac{dy}{dx}=2x-4[/tex]
By observation, this is equal to 0 when x=2
When x=2 the original equation becomes
[tex]y=(2)^2-4(2)-12[/tex]
[tex]y=-16[/tex]
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)
[tex]\Rightarrow[/tex] so its base is 8
The height of the triangle reaches from the midpoint of the base (-2, 0) and the vertex (2, -16)
[tex]\Rightarrow[/tex] so its height is 16
The area is [tex]\frac{1}{2}\times \text { base }\times \text { height }=\frac{1}{2}\times8\times16=64 \text{ sq units }[/tex]
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 = [tex]\frac{1}{2}(\text{base}\times \text{Height})[/tex]
= [tex]\frac{1}{2}(AB)(CD)[/tex]
= [tex]\frac{1}{2}(8\times 16)[/tex]
= 64 square units
Therefore, area of the triangle will be 64 square units.
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