What is the vertex of $y = x^{2} - 4 x - 15$ $y= x^2 - 4x- 15$ ?

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Answer 1 · 2016-01-02T21:39:52

(2 , -19 )

this is a quadratic function of the form $a x^{2} + b x + c$ $ax^2 + bx + c$

here a = 1 and b = -4
the x-coordinate of the vertex can be found using x = - b/2a

x = -( -4/2 )= 2
and y = $2^{2} - 4 \times 2 - 15$ $2^2 - 4 xx2 - 15$
y = 4 - 8 -15
y = -19

What is the vertex of y= x^2 - 4x- 15y=x2−4x−15 y= x^2 - 4x- 15?