To find the intercepts of the function h(t) = -5t^2 + 20t + 60, we need to set it equal to zero and solve for t.
1. Set h(t) = 0:
-5t^2 + 20t + 60 = 0
2. Solve the quadratic equation using the quadratic formula:
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
In this case, a = -5, b = 20, and c = 60. Plugging these values into the quadratic formula:
t = \frac{-20 \pm \sqrt{20^2 - 4(-5)(60)}}{2(-5)}
3. Simplify the equation:
t = \frac{-20 \pm \sqrt{400 + 1200}}{-10}
t = \frac{-20 \pm \sqrt{1600}}{-10}
t = \frac{-20 \pm 40}{-10}
4. Solve for t:
Case 1:
t = \frac{-20 + 40}{-10}
t = -2
Case 2:
t = \frac{-20 - 40}{-10}
t = 6
Therefore, the intercepts of the function h(t) are t = -2 and t = 6. This means that the graph of the function crosses the t-axis at t = -2 and t = 6.
To find the vertex of the function, we can use the formula for the x-coordinate of the vertex, which is given by:
x = -\frac{b}{2a}
In this case, a = -5 and b = 20. Plugging these values into the formula:
x = -\frac{20}{2(-5)}
x = -\frac{20}{-10}
x = 2
To find the y-coordinate of the vertex, we substitute the x-coordinate back into the function:
h(2) = -5(2)^2 + 20(2) + 60
h(2) = -20 + 40 + 60
h(2) = 80
Therefore, the vertex of the function h(t) is (2, 80).
Answer:
The intercepts of the function h(t) = -5t^2 + 20t + 60 are t = -2 and t = 6.
The vertex of the function h(t) is (2, 80).