What is the vertex form of #y=4x^2+19x - 5 #?

1 Answer
Binayaka C.
May 1, 2017

Vertex form of equation is #y = 4 (x+2.375)^2-27.5625#

Explanation:

#y=4x^2+19x-5 or y = 4 (x^2+19/4x) -5 or y = 4 (x^2+19/4x +19^2/8^2)-4*19^2/8^2 -5 # or

# y = 4 (x+19/8)^2-361/16 -5 or y = 4 (x+19/8)^2-441/16# or

#y = 4 (x+2.375)^2-27.5625# Comparing with standard vertex form of equation #y=a(x-h)^2 + k ;(h.k) # being vertex.

We get vertex at #(-2.375, -27.5625)#

Vertex form of equation is #y = 4 (x+2.375)^2-27.5625# [Ans]

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
969 views around the world
You can reuse this answer
Creative Commons License