How do you write $f(x)= -2x^2 + 5x - 8$ in vertex form?

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Answer 1 · 2016-09-07T05:59:39

$y=-2(x-5/4)^2-39/8$

Given -

$y=-2x^2+5x-8$

X-co-ordinate of the vertex

$x=(-b)/(2a)=(-(5))/(2 xx (-2))=(-5)/(-4)=5/4$
Y-coordinate of the vertex
At $x=5/4$

$y=-2(5/4)^2+5(5/4)-8=-39/8$

Vertex form of the equation is

$y=a(x-h)^2+k$

$a=-2$ coefficient of $x^2$
$h=5/4$ $x$ coordinate of the vertex
$k=-39/8$ $y$ coordinate of the vertex

Equation-

$y=-2(x-5/4)^2-39/8$

How do you write f(x)= -2x^2 + 5x - 8f(x)= -2x^2 + 5x - 8 in vertex form?