Answer:
a) The vertex form is [tex]f(x)=(x-9)^2+76[/tex]
b) The minimum value is 76.
Step-by-step explanation:
Given : function [tex]f(x)=x^2-18x+157[/tex]
We have to find the vertex form of f(a) and minimum value of f(x).
To write the vertex form.
Consider the given function [tex]f(x)=x^2-18x+157[/tex]
Vertex form a quadratic function [tex]f(x)=ax^2+bx+c[/tex] is [tex]f(x)=a(x-h)^2+k[/tex]where (h,k) is the vertex.
We write the square term in perfect square form that is in the form of [tex](a-b)^2=a^2+b^2-2ab[/tex]
Comparing we have a = x
-2ab = -18x
⇒ b = 9
Add and subtract [tex]b^2=81[/tex] in the given equation, we have,
[tex]f(x)=x^2-18x+81-81+157[/tex]
Simplify, we have,
[tex]f(x)=(x-9)^2+76[/tex]
Thus, The vertex form is [tex]f(x)=(x-9)^2+76[/tex]
b)
Minimum value of f(x) is at y value of vertex equation.
That is when x = 9 then value of function is [tex]f(x)=(9-9)^2+76=76[/tex]
Thus, The minimum value of given function [tex]f(x)=x^2-18x+157[/tex] is 76.
