What is the vertex form of $y = 8 x^{2} + 19 x + 12$ $y=8x^2 + 19x + 12$ ?

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Answer 1 · 2016-10-16T21:57:11

$y = 8 {(x - - \frac{19}{16})}^{2} + \frac{23}{32}$ $y = 8(x - -19/16)^2 + 23/32$

The equation is in the standard form, $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ where $a = 8, b = 19, and c = 12$ $a = 8, b = 19, and c = 12$

The x coordinate, h, of the vertex is:

$h = - \frac{b}{2 a}$ $h = -b/(2a)$

$h = - \frac{19}{2 (8)} = - \frac{19}{16}$ $h = -19/(2(8)) = -19/16$

To find the y coordinate, k, of the vertex, evaluate the function at the value of h:

$k = 8 (- \frac{19}{16}) (- \frac{19}{16}) + 19 (- \frac{19}{16}) + 12$ $k = 8(-19/16)(-19/16) + 19(-19/16) + 12$

$k = (\frac{1}{2}) (- 19) (- \frac{19}{16}) + 19 (- \frac{19}{16}) + 12$ $k = (1/2)(-19)(-19/16) + 19(-19/16) + 12$

$k = - \frac{19^{2}}{32} + 12$ $k = - 19^2/32 + 12$

$k = - \frac{361}{32} + 12$ $k = - 361/32 + 12$

$k = - \frac{361}{32} + \frac{384}{32}$ $k = - 361/32 + 384/32$

$k = \frac{23}{32}$ $k = 23/32$

The vertex form of the equation of a parabola is:

$y = a {(x - h)}^{2} + k$ $y = a(x - h)^2 + k$

Substitute our values into that form:

$y = 8 {(x - - \frac{19}{16})}^{2} + \frac{23}{32}$ $y = 8(x - -19/16)^2 + 23/32$

What is the vertex form of y=8x^2 + 19x + 12 y=8x2+19x+12y=8x^2 + 19x + 12 ?