Question

y = 240 - 10t - 5t^2 GRAPH

Answer 1

The graph of $y = 240 - 10t - 5t^2$ is a parabola opening downwards with vertex at $(-1, 245)$, axis of symmetry $t = -1$, t-intercepts at $(-8, 0)$ and $(6, 0)$, and y-intercept at $(0, 240)$.

Answer 2

The equation $y = 240 - 10t - 5t^2$ is a quadratic function, representing a parabola. Since the coefficient of $t^2$ is negative ($-5$), the parabola opens downwards. The vertex's t-coordinate is found using $t_v = -b/(2a)$. Here, $a = -5$ and $b = -10$. So, $t_v = -(-10) / (2 \times -5) = 10 / -10 = -1$. The vertex's y-coordinate is found by substituting $t_v$ into the equation: $y_v = 240 - 10(-1) - 5(-1)^2 = 240 + 10 - 5 = 245$. The vertex is at $(-1, 245)$. The axis of symmetry is the vertical line passing through the vertex, which is $t = -1$. To find the t-intercepts, set $y = 0$: $0 = 240 - 10t - 5t^2$. Dividing by $-5$ gives $t^2 + 2t - 48 = 0$. Factoring yields $(t+8)(t-6) = 0$, so $t = -8$ and $t = 6$. The t-intercepts are $(-8, 0)$ and $(6, 0)$. To find the y-intercept, set $t = 0$: $y = 240 - 10(0) - 5(0)^2 = 240$. The y-intercept is $(0, 240)$.