Ask Question
5 August, 14:29

Y=9x^2+9x-1 written in vertex form

+1
Answers (1)
  1. 5 August, 15:47
    0
    Y=9x^2+9x-1 is to be re-written in "vertex form," i. e., y-k = a (x-h) ^2, where a is a constant coefficient and (h, k) is the vertex.

    Y=9x^2+9x-1 can be re-written as y - 1 = 9 (x^2 + x).

    We can complete the square of x^2 + x:

    x^2 + 2 (1/2) + (1/2) ^2 - (1/2) ^2

    and this can be simplified as follows: (x+1/2) ^2 - 1/4

    Go back to y - 1 = 9 (x^2 + x) and subst. (x+1/2) ^2 - 1/4 for (x^2 + x):

    y - 1 = 9[ (x+1/2) ^2 - 1/4]

    Then the desired equation in vertex form is y-1 = 9[ (x+1/2) ^2 - (-1/4) ]

    or y - 1 = 9 [ (x+1/2) ^2 ] + 9/4, or y - 1 - (9/4) = 9 (x+1/2) ^2, or,

    finally, y - (13/4) = 9 (x+1/2) ^2. The vertex is at (1/2, 13/4).
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Y=9x^2+9x-1 written in vertex form ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers