Answer:
Given function: [tex]f(x)=3x^2-6x+5[/tex]
Vertex form: [tex]y=a(x-h)^2+k[/tex]
(where [tex](h, k)[/tex] is the vertex)
Expand vertex form:
[tex]y=ax^2-2ahx+ah^2+k[/tex]
Compare coefficients of given function with expanded vertex form
Comparing coefficient of [tex]x^2[/tex]:
[tex]3=a[/tex]
Comparing coefficient of [tex]x[/tex]:
[tex]\ \ \ \ \ -6=-2ah\\\implies-6=-2 \cdot 3h\\\implies -6=-6h\\\implies h=1[/tex]
Comparing constant:
[tex]\ \ \ \ \ \ 5=ah^2+k\\\implies5=3(1)^2+k \\\implies 5=3+k\\\implies k=2[/tex]
Therefore, the vertex is (1, 2)
As the leading coefficient is positive, the parabola will open upwards.
Additional plot points:
[tex]f(0)=3(0)^2-6(0)+5=5[/tex]
[tex]f(2)=3(2)^2-6(2)+5=5[/tex]
(0, 5) and (2, 5)
