What is the vertex of # y= –x^2 +12x – 4?

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What is the vertex of # y= –x^2 +12x – 4?

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Answer 1 · 2016-03-23T14:46:42

$Vertex \to (x, y) \to (6, 32)$ $color(blue)("Vertex"->(x,y)->(6,32)$

Explanation:

$General condition$ $color(blue)("General condition")$

Consider the standard form of $y = a x^{2} + b x + c)$ $y=ax^2+bx+c)$

Write this as $y = a (x^{2} + \frac{b}{a} x) + c$ $y=a(x^2+b/ax)+c$

$x_{vertex} = (- \frac{1}{2}) \times \frac{b}{a}$ $x_("vertex")=(-1/2)xxb/a$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Solving your question$ $color(blue)("Solving your question")$

In your case $a = - 1 and b = 12 \to x_{vertex} = (- \frac{1}{2}) \times \frac{12}{- 1} = + 6$ $a= -1 and b=12 ->x_("vertex")=(-1/2)xx12/(-1) = +6$

Substitute $x = 6 \to y_{vertex} = 32$ $x=6 -> y_("vertex")=32$

$Vertex \to (x, y) \to (6, 32)$ $color(blue)("Vertex"->(x,y)->(6,32))$

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What is the vertex of # y= –x^2 +12x – 4?

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