Question

Let ​ f(x)=x2−18x+157 ​.

Answer 1

Hello :
f(x)=x²−18x+157
      = x²-2(9)(x) +9²-9²+157
      = (x-9)² -9²+157
f(x) = (x-9)²+76....(the vertex form of f(x) )
Minimum value of f(x): y = 76

Answer 2

Answer:

a) The vertex form is $f(x)=(x-9)^2+76$

b) The minimum value is 76.

Step-by-step explanation:          

Given : function $f(x)=x^2-18x+157$

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 $f(x)=x^2-18x+157$

Vertex form a quadratic function $f(x)=ax^2+bx+c$ is $f(x)=a(x-h)^2+k$where (h,k) is the vertex.

We write the square term in perfect square form that is in the form of $(a-b)^2=a^2+b^2-2ab$

Comparing we have a = x

-2ab = -18x

⇒ b = 9

Add and subtract $b^2=81$ in the given equation, we have,

$f(x)=x^2-18x+81-81+157$  

Simplify, we have,

$f(x)=(x-9)^2+76$

Thus, The vertex form is $f(x)=(x-9)^2+76$

b)

Minimum value of f(x) is  at y value of vertex equation.

That is when x = 9  then  value of function is $f(x)=(9-9)^2+76=76$

Thus, The minimum value of given function $f(x)=x^2-18x+157$ is  76.