Answer:
vertex is (6 , -9)
points are (9,0) and (3,0)
Step-by-step explanation:
Given quadratic function \(f(x)=x^2-12x+27\)
We have to plot the given quadratic function.
Consider the Given quadratic function \(f(x)=x^2-12x+27\)
The general form of quadratic function is given \(f(x)=a(x-h)^2+k\)
Where, (h, k) is vertex , given \(h =\frac{-b}{2a}\) and \(k = f(h)\)
Thus, for given quadratic function \(f(x)=x^2-12x+27\)
a = 1 , b= -12 , c = 27
Thus,
\(h =\frac{-b}{2a}=\frac{12}{2}=6\)
\(k = f(h)\) that is f(12) = (6)^2 - 12× 6 +27 = 36 - 72 + 27 = - 9
Thus, given quadratic function \(f(x)=x^2-12x+27\) in standard form is \(f(x)=(x-6)^2-9\)
Thus, vertex is (6 , -9)
For second point put \(f(x)=0\) , we get,
\(f(x)=(x-6)^2-9=0\)
\(\Rightarrow (x-6)^2-9=0\)
\(\Rightarrow (x-6)^2=9\)
\(\Rightarrow (x-6)=\pm 3\)
\(\Rightarrow x= 6\pm 3\)
Thus, \(\Rightarrow x=6+3=9\) and \(\Rightarrow x=6-3=3\)
thus, points are (9,0) and (3,0)
Graph is attached below.
