Answer:
[tex]a+b+c=-\frac{7}{3}[/tex]
Step-by-step explanation:
Given:
The equation of the parabola with vertical axis of symmetry is given as:
[tex]y=ax^2+bx+c[/tex]
Vertex of the parabola is, [tex](h,k)=(5,3)[/tex]
Point on a parabola is (2,0).
The [tex]x[/tex] co-ordinate of the vertex is given as:
[tex]h=-\frac{b}{2a}\\5=-\frac{b}{2a}\\5\times 2a=-b\\10a=-b\\b=-10a----1[/tex]
Now, plug in the point (2,0) in the parabolic equation. This gives,
[tex]0=a(2)^2+b(2)+c\\0=4a+2(-10a)+c\\0=4a-20a+c\\c=20a-4a=16a------ 2[/tex]
Now, as vertex lies on the parabola, plug in (5,3) in the parabolic equation. This gives,
[tex]3=a(5)^2+b(5)+c\\3=25a+5b+c\\\textrm{But, b= -10a, c= 16a}\\3=25a+5(-10a)+16a\\3=25a-50a+16a\\3=-9a\\a=-\frac{3}{9}=-\frac{1}{3}----3[/tex]
So,
[tex]b=-10a=-10\times -\frac{1}{3}=\frac{10}{3}\\c=16a=16\times -\frac{1}{3}=-\frac{16}{3}[/tex]
Therefore, the sum of [tex]a+b+c[/tex] is:
[tex]=-\frac{1}{3}+\frac{10}{3}-\frac{16}{3}\\=\frac{-1+10-16}{3}\\=-\frac{7}{3}[/tex]
