Question 168083
Given the vertex (-5,-3) at points (1,2):
We can solve the eqn. using the Vertex Form in Solving Quadratic:
{{{f(x)=a(x-h)^2+k}}}
{{{f(x)=a(x-(-5))^2+(-3)}}}
{{{f(x)=a(x+5)^2-3}}}
{{{2=a(x^2+10x+25)-3}}}
{{{2=a(1^2+10*1+25)-3}}}
{{{2=36a-3}}}
{{{2+3=36a}}}
{{{cross(36)a/cross(36)=5/36}}} --->{{{a=5/36}}}
It follows the eqn:{{{f(x)=(5/36)x^2+(5/36)10x+(5/36)(25)}}}
{{{f(x)=(5/36)x^2+(50/36)x+(125/36)-3}}}
{{{f(x)=(5/36)x^2+(25/18)x+(125-108)/36}}}
{{{highlight(f(x)=(5/36)x^2+(25/18)x+(17/36))}}}, ANSWER
See graph:
{{{drawing(300,300,-20,10,-6,10,grid(1),graph(300,300,-20,10,-6,10,(5/36)x^2+(25/18)x+(17/36)),circle(-5,-3,.25),circle(1,2,.25)))}}} ----> See vertex (-5,-3) with points (1,2)
Thank you,
Jojo