What is the vertex of  y= -(2x-1)^2+x^2-x+3?

Apr 13, 2016

$\text{vertex } \to \left(x , y\right) \to \left(\frac{1}{2} , \frac{11}{4}\right)$

Explanation:

Multiply out the brackets giving:

$y = - \left(4 {x}^{2} - 4 x + 1\right) + {x}^{2} - x + 3$

Multiply everything inside the bracket by $\left(- 1\right)$ giving

$y = - 4 {x}^{2} + 4 x - 1 + {x}^{2} - x + 3$

$y = - 3 {x}^{2} + 3 x + 2$

Write as: $y = - 3 \left({x}^{2} + \frac{3}{- 3} x\right) + 2$

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

Consider the coefficient $- 1$ from $- x$ inside the brackets

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 1\right) = + \frac{1}{2}}$
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Substitute for x_("vertex") in the equation

color(brown)(y=-3x^2+3x+2" "->" " y=-3(color(blue)(1/2))^2+3(color(blue)(1/2) )+2

$\textcolor{b l u e}{{y}_{\text{vertex}} = 2 \frac{3}{4} = \frac{11}{4}}$
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color(blue)("vertex "->(x,y)->(1/2,11/4)#