Help please.
Find b and c so that y=13(x^2)+bx+c has a vertex (3,-2).
so b=______
and c=_____
Hello, lsnyder!
We're expected to know that the vertex is at: .$\displaystyle x \:=\:\frac{-b}{2a}$Find $\displaystyle b$ and $\displaystyle c$ so that $\displaystyle y\:=\:13x^2+bx+c$ has a vertex (3,-2).
So we have: .$\displaystyle \frac{-b}{2(13)} \:=\:3 \quad\Rightarrow\quad b \:=\:-78$
. . and the equation is: .$\displaystyle y \:=\:13x^2 - 78x + c$
Since (3,-2) is on the graph: .$\displaystyle -2 \:=\:13(3^2) - 78(3) + c \quad\Rightarrow\quad c \:=\:115$
Therefore: . $\displaystyle \boxed{b \:=\: -78}\quad\boxed{ c \:=\:115} $