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

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Answer 1 · 2015-12-04T03:23:00

$(-2/3,10/3)$

The vertex of a quadratic equation can be found through the vertex formula:

$(-b/(2a),f(-b/(2a)))$

The letters represent the coefficients in the standard form of a quadratic equation $ax^2+bx+c$ .

Here:
$a=-3$
$b=-4$

Find the $x$ -coordinate of the vertex.

$-b/(2a)=-(-4)/(2(-3))=-2/3$

The $y$ -coordinate is found by plugging $-2/3$ into the original equation.

$-3(-2/3)^2-4(-2/3)+2=-3(4/9)+8/3+2$
$=-4/3+8/3+6/3=10/3$

Thus, the vertex is located at the point $(-2/3,10/3)$ .

This can also be found through putting the quadratic into vertex form $y=a(x-h)^2+k$ by completing the square.

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

$y=-3(x^2+4/3x+color(blue)(4/9))+2+color(blue)(4/3)$

$y=-3(x+2/3)^2+10/3$

Again, the vertex is located at the point $(-2/3,10/3)$ .

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