## Algebra 1

The graph was made using graphing software. $f(x)= -2.5*x^2-x+3$ vertex: $x=-b/2a$ $x = -(-1)/2*(-2.5)$ $x = 1/-5 = -1/5$ $f(-.2)= -2.5*(-.2)^2-(-.2)+3$ $f(-.2) = -2.5*.04 +.2+3$ $f(-.2) = -.1+.2+3$ $f(-.2) = 3.1$ $(-.2, 3.1)$ Two other points on the curve: $x=0$ $f(0)= -2.5*(0)^2-(0)+3$ $f(0) = 0-0+3$ $f(0) = 3$ $(0,3)$ Since $(0,3)$ is .2 from the axis of symmetry, we also know $(.2, 3.1)$ is also on the curve.