## Algebra 1

$y=-.5x^2+4x+1$ vertex $x=-b/2a$ $x=-4/2*-.5$ $x=-4/-1 = 4$ $y=-.5x^2+4x+1$ $y=-.5*4^2+4*4+1$ $y=-.5*16+16+1$ $y=-8+17$ $y=9$ $(4,9)$ One other point on the line $x=0$ $y=-.5x^2+4x+1$ $y=-.5*0^2+4*0+1$ $y=-.5*0+0+1$ $y=0+0+1 = 1$ $(0,1)$ The vertex is labeled, and the axis of symmetry is the blue line.