The vertex is f(0) = 0^{2} + 3(0) + 5 = 5 The y-intercept is (0, 5). b^{2} – 4ac = 9 – 4(1)(5) = 9 – 20 = -11 The discriminate is negative, so there are no x-intercepts. You could also tell by seeing that the vertex is above the x-axis and a is positive, so the parabola points up. f(-1) = (-1)^{2} + 3(-1) + 5 = 1 – 3 + 5 = 3 f(-4) = (-4)^{2} + 3(-4) + 5 = 16 – 12 + 5 = 9 We need a few extra points, so we've chosen one from each side of the vertex. x | y | | | 0 | 5 | -1 | 3 | -4 | 9 | 3 | 5 | -2 | 3 | 1 | 9 |
(-3, 5), (-2, 3), and (1, 9) are symmetric to (0, 5), (-1, 3), and (-4, 9). |