What is the vertex form of $y = (9 x - 6) (3 x + 12) - 7 x^{2} + 5 x$ $y= (9x-6)(3x+12)-7x^2+5x$ ?

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Answer 1 · 2017-10-09T19:25:32

$y = 20 {(x - (- \frac{19}{8}))}^{2} - \frac{2957}{16}$ $y = 20(x-(-19/8))^2-2957/16$

Given: $y = (9 x - 6) (3 x + 12) - 7 x^{2} + 5 x$ $y= (9x-6)(3x+12)-7x^2+5x$

Perform the multiplication:

$y = 27 x^{2} + 90 x - 72 - 7 x^{2} + 5 x$ $y =27x^2 + 90x - 72 -7x^2+5x$

Combine like terms:

$y = 20 x^{2} + 95 x - 72$ $y =20x^2 + 95x - 72$

This is in the standard Cartesian form:

$y = a x^{2} + b x + c$ $y = ax^2+bx+c$

where $a = 20, b = 95, and c = - 72$ $a = 20, b = 95, and c = -72$

The general vertex form for a parabola of this type is:

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

We know that $a = 20$ $a = 20$ :

$y = 20 {(x - h)}^{2} + k$ $y = 20(x-h)^2+k$

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

$h = - \frac{95}{2 (20)}$ $h = -95/(2(20))$

$h = - \frac{19}{8}$ $h = -19/8$

$y = 20 {(x - (- \frac{19}{8}))}^{2} + k$ $y = 20(x-(-19/8))^2+k$

We know that:

$k = 20 {(- \frac{19}{8})}^{2} + 95 (- \frac{19}{8}) - 72$ $k = 20(-19/8)^2+95(-19/8)-72$

$k = - \frac{2957}{16}$ $k = -2957/16$

$y = 20 {(x - (- \frac{19}{8}))}^{2} - \frac{2957}{16}$ $y = 20(x-(-19/8))^2-2957/16$

What is the vertex form of y=(9x−6)(3x+12)−7x2+5x y= (9x-6)(3x+12)-7x^2+5x?