What is the vertex of $y = x^{2} + 6 x + 5$ $y=x^2+6x+5$ ?

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Answer 1 · 2016-01-05T14:57:53

( -3 , -4)

making use of the standard form of a trinomial ie. $a x^{2} + b x + c$ $ax^2 + bx + c$

for y = $x^{2} + 6 x + 5$ $x^2 + 6x + 5$

a = 1 , b = 6 and c = 5

the x-coordinate of the vertex = - (b/2a)

$\Rightarrow x = - \frac{6}{2}$ $rArr x = - 6/2$ = - 3

now substitute this value of x into the equation to obtain corresponding value of y.

$\Rightarrow y = {(- 3)}^{2} + 6 (- 3) + 5$ $rArr y = ( -3 )^2 + 6( -3 ) + 5$ = 9 - 18 + 5 = - 4
graph{x^2+6x+5 [-10, 10, -5, 5]}

What is the vertex of y=x2+6x+5y=x^2+6x+5?