The graph of y=ax^2+bx+c is given below, where a, b, and c are integers. Find a-b+c.
There is a falsity in the statement of this problem; a is definitely not an integer. It is even evident from the graph that a is around \(\frac{2}{3}\), since if you start with the vertex and move one unit to the right the y value increases by less than a unit; b and c are, however integers, b = -4 and c =7. You can find these values by writing the vertex form \(y=a{(x-h)}^{2}+k\) and plugging in 3 for h and 1 for k and expanding to get \(y = a({x}^{2}-6x+9) +1\). Finally, since the point (0, 7) is on the graph, we have 7 = 9a +1 , which yields \(a =\frac{2}{3}\). So \(b=\frac{2}{3}(-6)=-4\) and \(c=\frac{2}{3}(9) + 1= 7\). Finally, the parabola has equation
\(y=\frac{2}{3}{x}^{2}-4x+7.\)
