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

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Answer 1 · 2018-06-15T01:04:07

vertex is $(-2/5,-84/5)$

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

$y=3x^2+8x-16+2x^2-4x$

$y=5x^2+4x-16$

The vertex is given by $x=-b/(2a)$ where the quadratic equation is given by $y=ax^2+bx+c$

$x=-b/(2a) = -4/(2times5)=-4/10=-2/5$

Sub $x=-2/5$ into equation to get the y-value

$y=5(-2/5)^2+4(-2/5)-16$

$y=-84/5$

Therefore, your vertex is $(-2/5,-84/5)$

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