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

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Answer 1 · 2017-05-16T20:10:01

$Vertex (- \frac{6}{7}, \frac{823}{49})$ $"Vertex"(-6/7,823/49)$

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

$1-take derivative of the function with respect to x$ $"1-take derivative of the function with respect to x"$

$\frac{d y}{d x} = 2 (x - 3) \cdot 1 - 8 x - 1$ $(d y)/(d x)=2(x-3)*1-8x-1$

$1-equalize with zero and solve for x$ $"1-equalize with zero and solve for x"$

$2 (x - 3) - 8 x - 1 = 0$ $2(x-3)-8x-1=0$

$2 x - 6 - 8 x - 1 = 0$ $2x-6-8x-1=0$

$- 6 x - 7 = 0$ $-6x-7=0$

$- 6 x = 7$ $-6x=7$

$x = - \frac{6}{7}$ $x=-6/7$

$write x=-6/7 in the original equation and calculate for y$ $"write x=-6/7 in the original equation and calculate for y"$

$y = {(- \frac{6}{7} - 3)}^{2} - 4 {(- \frac{6}{7})}^{2} - (- \frac{6}{7}) + 4$ $y=(-6/7-3)^2-4(-6/7)^2-(-6/7)+4$

$y = {(- \frac{27}{7})}^{2} - 4 (\frac{36}{49}) + \frac{6}{7} + 4$ $y=(-27/7)^2-4(36/49)+6/7+4$

$y = \frac{729}{49} - \frac{144}{49} + \frac{34}{7}$ $y=729/49-144/49+34/7$

$y = \frac{585}{49} + \frac{34}{7}$ $y=585/49+34/7$

$y = \frac{585}{49} + \frac{238}{49}$ $y=585/49+238/49$

$y = \frac{823}{49}$ $y=823/49$

$y = 16.8$ $y=16.8$

What is the vertex of y= (x-3)^2-4x^2-x+4y=(x−3)2−4x2−x+4 y= (x-3)^2-4x^2-x+4?