What is the vertex form of $y = \frac{1}{2} x^{2} + \frac{3}{2} x - 4$ $y= 1/2x^2 + 3/2x -4$ ?

Question

1209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-30T20:10:05

$The vertex form is : y = \frac{1}{2} {(x + \frac{3}{2})}^{2} - \frac{41}{8}$ $"The vertex form is : "y=1/2(x+3/2)^2-41/8$

$The vertex form is formed as y= a {(x - h)}^{2} + k$ $"The vertex form is formed as y="a(x-h)^2+k$

$Where (h,k) is vertex coordinates$ $"Where (h,k) is vertex coordinates"$

$we should rearrange the given equation.$ $"we should rearrange the given equation."$

$y = \frac{1}{2} x^{2} + \frac{3}{2} x - 4$ $y=1/2x^2+3/2x-4$

$y = \frac{1}{2} x^{2} + \frac{3}{2} x + \frac{9}{8} - \frac{9}{8} - 4$ $y=1/2x^2+3/2xcolor(red)(+9/8-9/8)-4$

$y = \frac{1}{2} (x^{2} + 3 x + \frac{9}{4}) - \frac{9}{8} - 4$ $y=1/2(color(green)(x^2+3x+9/4))-9/8-4$

$x^{2} + 3 x + \frac{9}{4} = {(x + \frac{3}{2})}^{2}$ $color(green)(x^2+3x+9/4)=(x+3/2)^2$

$y = \frac{1}{2} {(x + \frac{3}{2})}^{2} - \frac{41}{8}$ $y=1/2(x+3/2)^2-41/8$

What is the vertex form of y=12x2+32x−4y= 1/2x^2 + 3/2x -4?