I am trying to solve the following exercise:
The graph of the fuction $y=-2x^2+bx+c$ passes through the point (1,0) and has as its vertex the point (3,S). What is the value of s?
Options: A -5_____ B 4______ C 8_____ D 18
Here's what I tried to do:
From the value of $a$(-2) I assumed that the concavity is down.
Since the first root of $x$ is 1 and the x-vertex is 3 I used the following formula to determine X2: $X2=-X1+2Xv$ (Xv being the x-vertex) . After that I got X2 = 5.
After finding X2 I determined $b$(6) and $c$(5)
The problem is that when I tried to determined y-vertex I got 5 as the result which is not in the options. Then for some reason(just guessing) I decided to change $c$ fom 5 to -5 so instead of having
$y=-2x^2+6x+5$
I had
$y=-2x^2+6x-5$
which gave me a final result of -5 which is in the options, but I refuse to believe that -5 is the y-vertex of the function because the concavity of the function is turned down and it has real roots on $x$ which makes it impossible to have a negative y-vertex.
What exactly am I doing wrong?
After you found $x_2=5$, you could say that
$y=-2(x-1)(x-5)=-2(x^2-6x+10)=-2x^2+12x-10$.
When $x=3$, you have that $y=-2(9)+12(3)-10=8$.
You could also work this by saying if $x=1$, $y=-2+b+c=0\implies b+c=2$.
Since $x=-\frac{b}{2a}$ at the vertex, $\frac{-b}{-4}=3\implies b=12$, so $c=-10$.