Question

determine the quadratic function of f whose graph is given. The vertex is (1,-3) and the y-intercept is -2

Answer 1

Hmmm the y-intercept is at -2?  what does that mean?  well, is where the graph "intercepts" or touches the y-axis, and when that happens, x = 0, so the point is really ( 0 , -2 ).

and we know where the vertex is at.  Let's assume a vertical parabola, in which case the squared variable is the "x".

$\bf \qquad \textit{parabola vertex form}\\\\ \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ vertex \begin{cases} h=1\\ k=-3 \end{cases}\implies y=a(x-1)^2-3 \\\\\\ \textit{now, we also know that } \begin{cases} x=0\\ y=-2 \end{cases}\implies -2=a(0-1)^2-3 \\\\\\ 1=a(-1)^2\implies \boxed{1=a}\qquad thus\implies \begin{cases} y=1(x-1)^2-3\\ \textit{or just}\\ y=(x-1)^2-3 \end{cases}$