Answer:
Domain = t ∈ ℝ or (-∞, ∞)
h = 6
Step by step explanation:
The given quadratic function is h = -16t^2 + 29t + 6.
To find the vertex, we can use the formula:
t = -b / 2a
where a = -16 and b = 29 are the coefficients of the t^2 and t terms, respectively.
Substituting the values of a and b, we get:
t = -29 / 2(-16) = 29 / 32
Substituting this value of t into the original equation, we can find the corresponding value of h:
h = -16(29/32)^2 + 29(29/32) + 6 ≈ 13.47
Therefore, the vertex of the parabola is (29/32, 13.47).
The domain of the function is all real numbers because there are no restrictions on the possible input values of t.
To find the range, we can consider that the coefficient of the t^2 term is negative, which means that the parabola opens downward. Therefore, the vertex is the maximum point of the parabola, and the range is all real numbers less than or equal to the y-coordinate of the vertex, which is approximately 13.47.
The x-axis is the set of values where h = 0. We can use the quadratic formula to find the roots of the equation:
-16t^2 + 29t + 6 = 0
The roots are approximately t = -0.15 and t = 1.92. Therefore, the parabola intersects the x-axis at t = -0.15 and t = 1.92.
The y-axis is the set of values where t = 0. Substituting t = 0 into the equation for h, we get:
h = -16(0)^2 + 29(0) + 6 = 6
Therefore, the parabola intersects the y-axis at h = 6.