How do you write $y = 4 x^{2} + 16 x + 23$ $y= 4x^2 + 16x + 23$ into vertex form?

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Answer 1 · 2015-05-13T03:12:08

$(- \frac{b}{2} a) = - \frac{16}{8} = - 2$ $(-b/2a) = -16/8 = -2$

$f (- \frac{b}{2} a) = f (- 2) = 16 - 32 + 23 = 7$ $f(-b/2a) = f(-2) = 16 - 32 + 23 = 7$

Factored form: $f (x) = 4 {(x + 2)}^{2} + 7$ $f(x) = 4(x + 2)^2 + 7$

Check:
Develop: $f (x) = 4 (x^{2} + 4 x + 4) + 7 = 4 x^{2} + 16 x + 23 .$ $f(x) = 4(x^2 + 4x + 4) + 7 = 4x^2 + 16x + 23.$ Correct

How do you write y= 4x^2 + 16x + 23y=4x2+16x+23y= 4x^2 + 16x + 23 into vertex form?