How do I use the vertex formula to determine the vertex of the graph for $y = - 4 x^{2} + 8 x$ $y = -4x^2 + 8x$ ?

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Answer 1 · 2015-05-01T14:36:52

The vertex of graph of $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

has $x$ $x$ -coordinate $x = - \frac{b}{2 a}$ $x=-b/(2a)$

So, the $x$ $x$ -coordinate of the vertex of the graph of $y = - 4 x^{2} + 8 x$ $y=-4x^2+8x$ is:

$x = - \frac{8}{2 (- 4)}$ $x= -(8)/(2(-4)$ , which we can evaluate (simplify, calculate)

$x = - \frac{8}{2 (- 4)} = - \frac{8}{- 8} = 1$ $x= -(8)/(2(-4)) = -8/-8 = 1$ ,

To find the $y$ $y$ -coordinate of the vertex, find the value of $y$ $y$ , when $x = 1$ $x=1$

$y = - 4 {(1)}^{2} + 8 (1) = - 4 + 8 = 4$ $y = -4(1)^2 +8(1) = -4 + 8 = 4$

The vertex is: $(1, 4)$ $(1, 4)$ .

How do I use the vertex formula to determine the vertex of the graph for y=−4x2+8xy = -4x^2 + 8x?