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