What is the vertex form of $y = 4 x^{2} + 12 x + 9$ $y=4x^2 + 12x + 9$ ?

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What is the vertex form of $y = 4 x^{2} + 12 x + 9$ $y=4x^2 + 12x + 9$ ?

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Answer 1 · 2016-05-17T15:38:17

$y = 4 {(x + 1.5)}^{2}$ $y=4(x+1.5)^2$

Explanation:

Given -

$y = 4 x^{2} + 12 x + 9$ $y=4x^2+12x+9$

Find the $x, y$ $x, y$ co-ordinates of the vertex

$x = \frac{- b}{2 a} = \frac{- 12}{2 \times 4} = \frac{- 12}{8} = - 1.5$ $x=(-b)/(2a)=(-12)/(2xx4)=(-12)/8=-1.5$

At $x = 1.5; y = 4 {(- 1.5)}^{2} + 12 (- 1.5) + 9$ $x=1.5; y= 4(-1.5)^2+12(-1.5)+9$

$y = 4 (2.25) - 18 + 9$ $y=4(2.25)-18+9$
$y = 9 - 18 + 9 = 0$ $y=9-18+9=0$

Equation in vertex form -

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$
$h$ $h$ = x- coordinate
$k$ $k$ = y- coordinate
$a$ $a$ = coefficient of $x^{2}$ $x^2$
Then -

$y = 4 {(x - (- 1.5))}^{2} + 0$ $y=4(x-(-1.5))^2+0$
$y = 4 {(x + 1.5)}^{2}$ $y=4(x+1.5)^2$

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What is the vertex form of $y = 4 x^{2} + 12 x + 9$ $y=4x^2 + 12x + 9$ ?

1 Answer

Explanation:

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