Answer:
Option 2 is correct.
The axis of symmetry is, x= 3
Step-by-step explanation:
Given: The zeros of the quadratic function f are 1 and 5.
If the zeroes are at x =1 and at x =5, then,
the factor equations were x- 1=0 and x-5 = 0.
Then. the factors were x -1 and x-5 .
Any factorable quadratic is going to have just the two factors, so these are must be them.
Then, the original quadratic was :
[tex](x-1)(x-5)= 0[/tex]
[tex]x(x-5) -1(x-5) =0[/tex] or
[tex]x^2-5x -x+5 =0[/tex] [by distributive property [tex]a\cdot (b+c) =a\cdot b+a\cdot c[/tex] ]
Combine like terms;
[tex]x^2-6x+5 =0[/tex] ......[1]
A quadratic equation is of the form: [tex]ax^2+bx+c =0[/tex]; where a, b, c are the coefficient ]
On comparing equation [1] with general equation we have;
the value of a = 1 , b = -6 and c =5
Axis of symmetry states that a parabola is a vertical line that divides the parabola into two congruent halves.
i,e [tex]x =-\frac{b}{2a}[/tex]
Then;
Axis of symmetry (x) = [tex]-\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3[/tex]
Therefore, the equation of the axis of symmetry is; x = 3
