What is the vertex of $y = 1/2 (3x + 4)^2 - 5$ ?

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Answer 1 · 2016-04-22T13:10:59

$color(blue)("Vertex"->(x,y)->(-4/3,-5)$

Consider the following:

Standard form $->y=ax^2+bx+c$

Vertex form $->y=a(x+b/(2a))^2+k+c$

Where $k=(-1)xxa(b/(2a))^2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(brown)("The given equation is not quite in vertex form")$

Write as: $" "y=3/2(x+4/3)^2-5" "$ Now it is!

$color(blue)(x_("vertex")=color(brown)((-1)xxb/(2a))color(green)( = (-1)xx4/3 )= -4/3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)(y_("vertex")=color(brown)(k+c) =-5$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("Vertex"->(x,y)->(-4/3,-5)$

Tony B Tony B

What is the vertex of y = 1/2 (3x + 4)^2 - 5 y = 1/2 (3x + 4)^2 - 5 ?