Suppose that we have an equation $y=ax^2+bx+c$ whose graph is a parabola with vertex $(3,2)$, vertical axis of symmetry, and contains the point $(1,0)$. What is $(a, b, c)$?

Guest Jun 30, 2019

#1**+5 **

The equation of a vertical parabola with a vertex of (3, 2) in vertex form is

y = A(x - 3)^{2} + 2

We know that (1, 0) is a solution to the equation. We can use that to find A

0 = A(1 - 3)^{2} + 2

0 = A( -2 )^{2} + 2

0 = 4A + 2

-4A = 2

A = -\(\frac12\)

So the equation of the parabola is

y = -\(\frac12\)(x - 3)^{2} + 2 Now we just have to get this equation into the form ax^{2} + bx + c

y = -\(\frac12\)(x - 3)(x - 3) + 2

y = -\(\frac12\)(x^{2} - 6x + 9) + 2

y = -\(\frac12\)x^{2} + 3x - \(\frac92\) + 2

y = -\(\frac12\)x^{2} + 3x - \(\frac{5}{2}\)

Now we can see that (a, b, c) = ( -\(\frac12\), 3, -\(\frac{5}{2}\) )

hectictar Jun 30, 2019