What is the vertex of $y=4/3x^2 - 2x - 3$ ?

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Answer 1 · 2017-11-16T02:05:30

$Vertex(3/4,-15/4)$

In this form of the Parabola equation, i.e.:

$ax^2+bx+c$

the vertex has coordinates of:

$x=-b/(2a)$ and $y=f(-b/(2a))$

In this problem:

$a=4/3$ and $b=-2$ and $c=-3$

$x$ -coordinate of the vertex = $(-(-2))/(2(4/3))=2/(8/3)=2*(3/8)=3/4$

$y$ -coordinate of the vertex can be found by plugging in the value of the $x$ -coordinate into the equation of the Parabola.

$y=(4/3)(3/4)^2-2(3/4)-3$

$y=(4/3)(9/16)-(3/2)-3$

$y=3/4-3/2-3$

$y=(3-6-12)/4=-15/4$

What is the vertex of y=4/3x^2 - 2x - 3 y=4/3x^2 - 2x - 3 ?