Find the distance between the vertices of the parabolas y = − 1/2x^2 + 4x and y = 2x^2 − 4x − 1.?
I need help with the process of the question. If anyone would be willing to help
More Questions From This User See All
I need help with the process of the question. If anyone would be willing to help
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
y
= (-1/2)x^2 + 4x
= (-1/2)(x^2 - 8x)
= (-1/2)(x - 4)^2 - (-1/2)(4^2)
= (-1/2)(x - 4)^2 + 8
Vertex is (4, 8).
y
= 2x^2 - 4x - 1
= 2(x^2 - 2x) - 1
= 2(x - 1)^2 - (2)(1^2) - 1
= 2(x - 1)^2 - 2 - 1
= 2(x - 1)^2 - 3
Vertex is (1, -3).
Distance between vertices
= sqrt((-3 - 8)^2 + (1 - 4)^2)
= sqrt((-11)^2 + (-3)^2)
= sqrt(121 + 9)
= sqrt(130)