we have the function
[tex]f(x)=2x^2+16x+31[/tex]
This is a vertical parabola open upward
The vertex is a minimum
Convert the given equation into a vertex form
[tex]f(x)=a(x-h)^2+k[/tex]
where
(h,k) is the vertex
so
step 1
factor the leading coefficient 2
[tex]\begin{gathered} f(x)=(2x^2+16x)+31 \\ f(x)=2(x^2+8x)+31 \end{gathered}[/tex]
step 2
Complete the square
[tex]\begin{gathered} f(x)=2(x^2+8x+4^2-4^2)+31 \\ f(x)=2(x^2+8x+4^2)+31-32 \\ f(x)=2(x^2+8x+4^2)-1 \end{gathered}[/tex]
step 3
Rewrite as a perfect square
[tex]f(x)=2(x+4)^2-1[/tex]
The vertex is the point (-4,-1)
Find out the y-intercept (value of f(x) when the value of x=0)
In the original expression
[tex]\begin{gathered} f(x)=2(0)^2+16(0)+31 \\ f(x)=31 \end{gathered}[/tex]
The y-intercept is the point (0,31)
using a graphing tool
Find out another point
For x=-2
substitute
[tex]\begin{gathered} f(x)=2(-2+4)^2-1 \\ f(x)=2(2)^2-1 \\ f(x)=7 \end{gathered}[/tex]
so
the other point is (-2,7)
