What is the vertex of $y= -5x^2 + 8x - 13$ ?

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Answer 1 · 2018-07-09T18:21:43

$x=4/5,y=-49/5$

We get

$y'(x)=-10x+8$ (to find the vertex coordinates)
$y'(x)=0$ if $x=4/5$

now we compute the $y$ coordinate:

$f(4/5)=-5*(16)/(25)+8*4/5-13=(32-16-65)/5=-49/5$

What is the vertex of y= -5x^2 + 8x - 13 y= -5x^2 + 8x - 13 ?