How do you find the vertex of $y = 4x^2 – 6x – 3$ ?

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Answer 1 · 2017-07-07T08:46:08

The vertex is the point at which the gradient is equal to $0$ .

The vertex is the point $(3/4,-21/4)$

The vertex is the point at which the first derivative (the gradient) is equal to $0$ .

$y=4x^2-6x-3$
$y'=8x-6$

$8x-6=0 -> x=6/8=3/4$

Substitute this x-value back into the original equation to find the y coordinate of the vertex:

$y=4x^2-6x-3=4(3/4)^2-6(3/4)-3$

$y= 36/16-18/4-3=9/4-18/4-12/4=-21/4$

So the vertex is the point $(3/4,-21/4)$

How do you find the vertex of y = 4x^2 – 6x – 3 y = 4x^2 – 6x – 3?