I can't solve the last exercises in a worksheet of Pre-Calculus problems. It says:
Quadratic function $f(x)=ax^2+bx+c$ determines a parabola that passes through points $(0, 2)$ and $(4, 2)$, and its vertex has coordinates $(x_v, 0)$.
a) Calculate coordinate $x_v$ of parabola's vertex.
b) Calculate $a, b$ and $c$ coefficients.
How can I get parabola's equation with this information and find what is requested?
I would appreciate any help. Thanks in advance.
Best Answer
Since $f(0)=c$ and we are given $f(0)=2$, we see immediately that $c=2.$
Furthermore, the equation in vertex form is $f(x)=a(x-x_v)^2+k$,
and since we are given $f(x_v)=0$, we see that $k=0,$ i.e., $f(x)=a(x-x_v)^2$.
From $a(x-x_v)^2=ax^2+bx+2$ we see that $ax_v^2 = 2$ and $-2ax_v=b.$
Since $f(4)=f(0)=2$, $(4-x_v)^2=x_v^2$, which means $x_v=2$. Thus $a=\frac12$ and $b=-2.$