Thomas' Calculus 13th Edition

The first part of the graph is the parabolic function $4-x^2$ until the point (1,3), which cuts the x-axis at (-2,0). The second part of the graph is the parabolic function $x^2 +2x$ from the point (1,3) on through (2,8).
$F(1) = 4 - 1^2$ $F(1) = 4 - 1$ $F(1) = 3$ The first part will cut the x-axis at: $4-x^2 = 0$ $x^2 = 4$ $x = \sqrt4$ or $x = -\sqrt4$ $x = 2$ or $x = -2$ So the first part of the graph will be the parabolic function $4-x^2$ until the point (1,3), which cuts the x-axis at (-2,0). $F(1) = 1^2 + 2 \times 1$ $F(1) = 1 + 2$ $F(1) = 3$ $F(2) = 2^2 + 2 \times 2$ $F(2) = 4 + 4$ $F(2) = 8$ So the second part of the graph will be the parabolic function $x^2 +2x$ from the point (1,3) on through (2,8).