What is the vertex form of $y=4x^2-49x - 5$ ?

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Answer 1 · 2015-12-05T01:08:13

vertex $((-49)/8, 445 3/16)$

Given -
$y= 4x^2 -49x-5$

If the quadratic equation is in the form
$ax^2+bx+c$ then its vertex is given by $(-b)/(2a)$

x $= (-49)/(2 xx 4)=(-49)/8$

At $x=(-49)/8$

$y=4((-49)/8)-49((-49)/8)-5=445 3/16$

vertex $((-49)/8, 445 3/16)$

What is the vertex form of y=4x^2-49x - 5 y=4x^2-49x - 5 ?