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

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Lucy Share
Jun 15, 2018

vertex is $\left(- \frac{2}{5} , - \frac{84}{5}\right)$

Explanation:

$y = \left(x + 4\right) \left(3 x - 4\right) + 2 {x}^{2} - 4 x$

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

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

The vertex is given by $x = - \frac{b}{2 a}$ where the quadratic equation is given by $y = a {x}^{2} + b x + c$

$x = - \frac{b}{2 a} = - \frac{4}{2 \times 5} = - \frac{4}{10} = - \frac{2}{5}$

Sub $x = - \frac{2}{5}$ into equation to get the y-value

$y = 5 {\left(- \frac{2}{5}\right)}^{2} + 4 \left(- \frac{2}{5}\right) - 16$

$y = - \frac{84}{5}$

Therefore, your vertex is $\left(- \frac{2}{5} , - \frac{84}{5}\right)$

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