The Question is: write the equation of the quadratic function with zeros at $(0,0)$ and $(6,0)$ with $f(5) = -15$.
So, I know how to get the equation from the zeros, but I am confused with what I am supposed to do with "$f(5) = -15$".
Does it matter? I can do my own research, but have no idea what that is. ):
Thank you for any help :)
Sven Dysthe
You are given that the roots are $(0,0)$ and $(6,0)$. This means that if if $f(x) = ax^2 + bx + c $, then you can write $f(0) = 0 $ and $f(6)=0$ you are also given $f(5)=-15$.
We have $f(0) = c = 0$. Hence, can write $f(x) = ax^2 + bx$. We still need to find $a$ and $b$. Since $f(6) = 0,$ we have $36a + 6b = 0 $. And since $f(5) = -15,$ then $25a + 5b = - 15 $ which means (after simplifying) $5a + b = -3$. Solve this for $b$ to obtain $b = -3 - 5a $. Substitute this into $36a + 6b = 0$, to obtain
$$ 36 a + 6(-3-5a) = 0 \implies 36a -18 - 30a = 0 \implies 6a = 18 \implies a = 3$$
and putting this into $b = -3-5a$, we obtain that $b = -3 -5(3) = - 18$. In other words, $\boxed{ f(x) = 3x^2 - 18x }$