How do you write #y=-2(x+1)^2+3#?

1 Answer
Jun 22, 2018

See a solution process below:

Explanation:

First, use this rule for quadratics to expand the term within the parenthesis:

#(color(red)(a) + color(blue)(b))^2 = (color(red)(a) + color(blue)(b))(color(red)(a) + color(blue)(b)) = color(red)(a)^2 + 2color(red)(a)color(blue)(b) + color(blue)(b)^2#

#y = -2(color(red)(x) + color(blue)(1))^2 + 1#

#y = -2(color(red)(x) + color(blue)(1))(color(red)(x) + color(blue)(1)) + 1#

#y = -2(color(red)(x)^2 + (2 * color(red)(x) * 1) + color(blue)(1)^2) + 1#

#y = -2(x^2 + 2x + 1) + 1#

Next, we can expand the term within the parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#y = color(red)(-2)(x^2 + 2x + 1) + 1#

#y = (color(red)(-2) xx x^2) + (color(red)(-2) xx 2x) + (color(red)(-2) xx 1) + 1#

#y = -2x^2 + (-4x) + (-2)) + 1#

#y = -2x^2 - 4x - 2 + 1#

Now, we can combine the common terms, for this problem the constants:

#y = -2x^2 - 4x - 1#