What is the vertex form of $y = 4 x^{2} + 8 x - 19$ $y=4x^2 + 8x − 19$ ?

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What is the vertex form of $y = 4 x^{2} + 8 x - 19$ $y=4x^2 + 8x − 19$ ?

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Answer 1 · 2017-11-23T18:14:46

$(- 1, - 23)$ $(-1,-23)$

Explanation:

The vertex equation is:
$x_{v} = \frac{- b}{2 a}$ $x_v=(-b)/(2a)$

for these function,
$x_{v} = \frac{- 8}{2 \cdot 4} = - 1$ $x_v=(-8)/(2*4)=-1$

now we substitute $x$ $x$ by $- 1$ $-1$ in the function equation,
$f (- 1) = 4 \cdot {(- 1)}^{2} + 8 \cdot (- 1) - 19 = - 23$ $f(-1)=4·(-1)^2+8·(-1)-19=-23$

so the vertex is point $(- 1, - 23)$ $(-1,-23)$ .

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What is the vertex form of $y = 4 x^{2} + 8 x - 19$ $y=4x^2 + 8x − 19$ ?

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Explanation:

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