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

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Answer 1 · 2017-05-01T10:27:32

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

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

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