Graph the parabola
[tex]\begin{gathered} y=x^2-10x+27 \\ f(x)=ax^2+bx+c \end{gathered}[/tex]
In order to find the vertex (h,k), we can use this formula
[tex]\begin{gathered} h=\frac{-b}{2a} \\ k=f(h) \end{gathered}[/tex]
where,
a = 1
b = -10
c = 27
then, the vertex (h,k) is
[tex]\begin{gathered} h=-\frac{-10}{2\cdot1}=\frac{10}{2}=5 \\ k=f(5)=5^2-10\cdot5+27=25-50+27=2 \end{gathered}[/tex]
Therefore, vertex is the point (h,k) = (5,2)
Now, we just need two points to the left and two points to the right of this point
for example, when x = 3, then y = 6
[tex]f(3)=3^2-10\cdot\: 3+27=6[/tex]
when x = 4, then y = 3
[tex]f(4)=4^2-10\cdot\: 4+27=3[/tex]
when x = 6, then y = 3
[tex]f(6)=6^2-10\cdot\: 6+27=3[/tex]
when x = 7, then y = 6
[tex]f(7)=7^2-10\cdot\: 7+27=6[/tex]
Thus, the set of 5 points is the following:
[tex](3,6),(4,3),(5,2),(6,3),(7,6)[/tex]
