How do you find the vertex and intercepts for $y = 2 x^{2} - 16 x + 27$ $y=2x^2-16x+27$ ?

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Answer 1 · 2016-05-07T21:16:34

$y = 2 {(x - 4)}^{2} - 5$ $y = 2(x - 4)^2 - 5" "$ Vertex is at $(4, - 5)$ $(4, -5)$
$y -$ $y-$ intercept is at 27

$y = 2 x^{2} - 16 x + 27$ $y = 2x^2 - 16x + 27$
$y = 2 (x^{2} - 8 x + \frac{27}{2})$ $y = 2(x^2 - 8x + 27/2) " "$ now complete the square

$y = 2 [x^{2} - 8 x + 16 - 16 + \frac{27}{2}] \Rightarrow$ $y = 2[color(blue)(x^2 - 8x + 16) -16 +27/2] " " rArr$ +16-16 = 0

$y = 2 [{(x - 4)}^{2} - 16 + \frac{27}{2}]$ $y = 2[color(blue)((x - 4)^2) -16 +27/2] " "$ $- 16 + \frac{27}{2} = - \frac{5}{2}$ $-16+27/2 = -5/2$

$y = 2 {(x - 4)}^{2} - 2 \times \frac{5}{2}$ $y = 2(x-4)^2 - 2xx5/2$

$y = 2 {(x - 4)}^{2} - 5$ $y = 2(x - 4)^2 - 5$

How do you find the vertex and intercepts for y=2x2−16x+27y=2x^2-16x+27?