+10 Find area of the isosceles triangle formed by the vertex and the x-intercepts of parabola y=x^2-4x+21.
+128460 y = x^2 - 4x + 21
x coordinate of the vertex 4 / (2) = 2
y coordinate of the vertex = 2^2 - 4(2) + 21 = 17
The verxtex is ( 2,17).....above the x axis...since this parabola turns upward, there aren't any x intercepts...
However...if you mean y = -x^2 -4x + 21
The x coordinate of the vertex is 4/ (2 * -1) = -2
And the y coordinate of the vertex is - (-2)^2 - 4(-2) + 21 = -4 + 8 + 21 = 25....this is the height of the triangle
We can find the x intercepts as
-x^2 -4x + 21 = 0 multiply through by -1
x^2 + 4x - 21 = 0 tactor
(x + 7) ( x - 3) = 0
Set each factor to 0 and solv for x and we get that x= -7 and x = 3
The base of the triangle = ( 3 - - 7) =10
So....the area is (1/2) (base)(height) = (1/2) (10)(25) = 125
