What is the vertex of $y= -3x^2-x-3-(x-3)^2$ ?

Question

1688 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T08:02:35

the vertex is at $(-0.875, 9.0625)$

$y = −3x^2 −x −3 −(x −3)^2$

Simplify the RHS
$y = -3x^2 -x -3 - x^2 - 6x +9$
$y = -4x^2 -7x +6$

The general quadratic form is $y = ax2 + bx + c$
The vertex can be found at $(h,k)$
where $h = -b/2a$

Substitute in what we know
$h = -(-7)/(2*-4) = -7/8 = -0.875$

Substititue the value of $h$ for $x$ in the original equation
$y = -4(-7/8)^2 -7(-7/8) +6 = 9.0625$

the vertex is at $(-0.875, 9.0625)$

What is the vertex of y= -3x^2-x-3-(x-3)^2 y= -3x^2-x-3-(x-3)^2?