What is the vertex of $y= 4(x+2)^2-2x^2-3x-1$ ?

Question

1527 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-20T05:44:16

Vertex : $(-13/4 , -49/8)$

Vertex form: $y = 2(x+13/4)^2 -49/8$

Step 1: Expand/multiply the function so it can be int eh standard form of $y = ax^2 + bc+c$

Given
$y = 4(x+2)^2 -2x -3x -1$

$= 4(x+2)(x+2)-2x^2 -3x-1$

$= 4(x^2 +2x+2x+4) -2x^2 -3x-1$

$=4(x^2 +4x+4) = 2x^2 -3x -1$

$=4x^2 +16 x +16 -2x^2 -3x -1$

$=2x^2 + 13x+15$

$a = 2, " " " b= 13, " " " c= 15$

The formula for vertex is $(-b/(2a), f(-b/(2a)))$

$x_(vertex) = -b/(2a) = h$

$x_(vertex) = (-13)/(2*2) = -13/4$

$y_(vertex) = f(-b/(2a)) = k$

$f(-13/4) = 2(-13/4)^2 +13(-13/4)+15$

$= 2(169/16)-169/4 +15 = -49/8$

Vertex form: $y = a(x-h)^2 + k$

$y = 2(x+13/4)^2 -49/8$

What is the vertex of y= 4(x+2)^2-2x^2-3x-1 y= 4(x+2)^2-2x^2-3x-1?