What is the vertex form of $y= (3-x)(3x-1)+11$ ?

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Answer 1 · 2016-05-02T23:03:40

$y = -3(x-5/3)^2+49/3$

The vertex form of a quadratic equation is $y = a(x-h)^2+k$ . In this form, we can see that the vertex is $(h, k)$ .

To put the equation in vertex form, first we'll expand the equation, and then use a process called completing the square.

$y=(3-x)(3x-1)+11$

$=> y = -3x^2+9x+x-3+11$

$=> y = -3x^2+10x+8$

$=> y = -3(x^2-10/3x)+8$

$=> y = -3(x^2-10/3x+(5/3)^2-(5/3)^2)+8$

$=> y = -3(x^2-10/3x+25/9)+(-3)(-25/9)+8$

$=> y = -3(x-5/3)^2+49/3$

So, the vertex form is $y = -3(x-5/3)^2+49/3$ and the vertex is $(5/3,49/3)$

What is the vertex form of y= (3-x)(3x-1)+11 y= (3-x)(3x-1)+11 ?