Question

Find $a+b+c$ if the graph of the equation $y=ax^2+bx+c$ is a parabola with vertex $(5,3)$, vertical axis of symmetry, and contains the point $(2,0)$.

Answer 1

Answer:

$a+b+c=-\frac{7}{3}$

Step-by-step explanation:

Given:

The equation of the parabola with vertical axis of symmetry is given as:

$y=ax^2+bx+c$

Vertex of the parabola is, $(h,k)=(5,3)$

Point on a parabola is (2,0).

The $x$ co-ordinate of the vertex is given as:

$h=-\frac{b}{2a}\\5=-\frac{b}{2a}\\5\times 2a=-b\\10a=-b\\b=-10a----1$

Now, plug in the point (2,0) in the parabolic equation. This gives,

$0=a(2)^2+b(2)+c\\0=4a+2(-10a)+c\\0=4a-20a+c\\c=20a-4a=16a------ 2$

Now, as vertex lies on the parabola, plug in (5,3) in the parabolic equation. This gives,

$3=a(5)^2+b(5)+c\\3=25a+5b+c\\\textrm{But, b= -10a, c= 16a}\\3=25a+5(-10a)+16a\\3=25a-50a+16a\\3=-9a\\a=-\frac{3}{9}=-\frac{1}{3}----3$

So,

$b=-10a=-10\times -\frac{1}{3}=\frac{10}{3}\\c=16a=16\times -\frac{1}{3}=-\frac{16}{3}$

Therefore, the sum of $a+b+c$ is:

$=-\frac{1}{3}+\frac{10}{3}-\frac{16}{3}\\=\frac{-1+10-16}{3}\\=-\frac{7}{3}$