If f(x) = 3x^2-10x+2 find the vertex of y=f(x+4)-5

1 answer:

3 0

$3 {(4 + x)}^{2} - 10(x + 4) + 2 - 5$
so
$3( {x}^{2} + 8x + 16) - 10x - 40 + 3$
so
$3 {x}^{2} + 14x + 11$
take differential calculus
$6x + 14 = 0$
you will find x then y

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