Question

what is the vertex of y=x^2-11x+28

Answer 1

http://mathway.com/graph_image.aspx?p=ySMB01xSMB072-11x+28_(4,0)_(7,0)_(0,28)?p=300?p=300?p=False?p=False?p=False?p=False
x=4,7, y=28

Answer 2

The vertex can be found by taking the midpoint of the two zeros (4 and 7) and pluging that number into the equation.
So, the midpoint is x = 5.5
y = (5.5)^2 - 11(5.5) + 28
y = 30.25 - 60.5 + 28
y = -2.25
The vertex is (5.5, -2.25)

Answer 3

\[vertex=(p,q)\]
\[p=\frac{-b}{2a}....q=c-ap^2\]
\[y(x)=ax^2+bx+c\]\[a=1.........b=-11........c=28\]\[p=\frac{-(-11)}{2}=\frac{11}{2}....................q=28-(1)(11/2)^2=\frac{-9}{4}\]\[vertex (\frac{11}{2},\frac{-9}{4})\]

Answer 4

senoins method works but there is the possibility that x intercepts will not exist.