What is the the vertex of $y = (x+6)(x+4) -x+12$ ?

Question

2354 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-23T11:16:47

$y_{min} = 63/4$ at $x = - 9/2$

$y = (x+6)(x+4) -x+12$
$y = x^2 + 10x + 24 -x+12$
$y = x^2 + 9x + 36$
$y = (x + 9/2)^2 - 81/4 + 36$
$y = (x + 9/2)^2 + 63/4$

$y_{min} = 63/4$ at $x = - 9/2$

Desmos Desmos

Answer 2 · 2016-06-23T12:09:38

The vertex is $(-9/2;63/4)$

let's rewrite the equation in the equivalent form:

$y=x^2+4x+6x+24-x+12$

$y=x^2+9x+36$

Then let's find the vertex coordinates by the following:

$x_V=-b/(2a)$

where a=1; b=9

so

$x_V=-9/2$

and

$y_V=f(-9/2)$

that's

$y=(-9/2)^2+9(-9/2)+36$

$y=81/4-81/2+36$

$y=(81-162+144)/4$

$y=63/4$

What is the the vertex of y = (x+6)(x+4) -x+12y = (x+6)(x+4) -x+12?