Answer:
None of the options is the answer to the question
Step-by-step explanation:
we know that
The equation of a vertical parabola into vertex form is equal to
[tex]f(x)=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
case A) we have
[tex]f(x)=x^{2}+4x-11[/tex]
In this case the x-coordinate of the vertex will be negative
therefore
case A is not the solution
case B) we have
[tex]f(x)=-2x^{2}+16x-35[/tex]
This case is a vertical parabola open downward (the vertex is a maximum)
The vertex is the point [tex](4,-3)[/tex] but is not a minimum
see the attached figure
therefore
case B is not the solution
case C) we have
[tex]f(x)=x^{2}-4x+5[/tex]
Convert into vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-5=x^{2}-4x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-5+4=(x^{2}-4x+4)[/tex]
[tex]f(x)-1=(x^{2}-4x+4)[/tex]
Rewrite as perfect squares
[tex]f(x)-1=(x-2)^{2}[/tex]
[tex]f(x)=(x-2)^{2}+1[/tex] --------> vertex form
The vertex is the point [tex](2,1)[/tex]
therefore
case C is not the solution
case D) we have
[tex]f(x)=2x^{2}-16x+35[/tex]
Convert into vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-35=2x^{2}-16x[/tex]
Factor the leading coefficient
[tex]f(x)-35=2(x^{2}-8x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-35+32=2(x^{2}-8x+16)[/tex]
[tex]f(x)-3=2(x^{2}-8x+16)[/tex]
Rewrite as perfect squares
[tex]f(x)-3=2(x-4)^{2}[/tex]
[tex]f(x)=2(x-4)^{2}+3[/tex] --------> vertex form
The vertex is the point [tex](4,3)[/tex]
therefore
case D is not the solution
The answer to the question will be the function
[tex]f(x)=2x^{2}-16x+29[/tex]
