Question

Find the vertex of {x}^{2} - 16x + 5" alt="4 {x}^{2} - 16x + 5" align="absmiddle" class="latex-formula">

Answer 1

To find the x coordinate of the vertex,

x = $\frac{-b}{2a}$

In this case, a = 4 and b = -6. Sub these values into the equation,

x = $\frac{-(-6)}{2(4)}$
   = 2

Sub x = 2 back into the original equation,

$4(2)^{2} -16(2)+5$
= -11

Therefore, your vertex is V(2, -11)

Hope this helped!