A frog jumps from a log to the ground below. It’s height, h, in centimeters as a function of time, t, in seconds, can be modeled by h(t) = -490t^2 + 75t + 12
what is the h intercept and the t intercept?
and what is the domain and range of this function and what does it mean in this context.
To find the h-intercept, we need to find the value of h when t = 0. We can substitute t = 0 into the equation h(t) = -490t^2 + 75t + 12:
h(0) = -490(0)^2 +75(0) + 12
h(0) = 0 + 0 + 12
h(0) = 12
Therefore, the h-intercept is h = 12.
To find the t-intercept, we need to find the value of t when h = 0. We can set h(t) = 0 and solve for t:
-490t^2 + 75t + 12 = 0
We can use the quadratic formula to solve this equation:
t = (-b ± √(b^2 - 4ac))/(2a)
where a = -490, b = 75, and c = 12.
Using the quadratic formula, we can calculate t:
t = (-75 ± √(75^2 - 4 * -490 * 12))/(2 * -490)
After solving the equation, we find two possible values for t: t ≈ 0.08 and t ≈ 0.18.
Therefore, the t-intercepts are t ≈ 0.08 and t ≈ 0.18.
Now, let's consider the domain and range of the function h(t) = -490t^2 + 75t + 12.
The domain of the function is the set of all valid inputs, which in this case is the set of all real numbers, as there are no restrictions on t.
The range of the function is the set of all possible outputs that the function can produce. Since the coefficient of the t^2 term is negative (-490), the graph of the function is a downward-opening parabola. Therefore, the range is all real numbers less than or equal to the maximum value of h(t).
To find the maximum value of h(t), we can find the vertex of the parabola. The vertex of a parabola in the form ax^2 + bx + c is given by the coordinates (-b/(2a), f(-b/(2a))), where f(x) represents the function.
In this case, the vertex of the parabola -490t^2 + 75t + 12 is given by:
t = -b/(2a) = -75/(2*-490) ≈ 0.077
Substituting t ≈ 0.077 into the function h(t), we get:
h ≈ -490(0.077)^2 + 75(0.077) + 12 ≈ 4.75
Therefore, the maximum value of h(t) is approximately 4.75.
Hence, the range of the function h(t) = -490t^2 + 75t + 12 is all real numbers less than or equal to 4.75.
In the context of the frog jumping from a log to the ground, the h-intercept represents the initial height of the frog above the ground when the time is 0 seconds. In this case, it is 12 centimeters.
The t-intercepts represent the points in time when the frog is at the same height as the ground, and they are approximately 0.08 seconds and 0.18 seconds.
The domain of the function represents all possible values of time during the frog's jump. Since there are no restrictions, the domain is all real numbers.
The range of the function represents all possible heights the frog can reach during its jump. In this case, it is all real numbers less than or equal to 4.75 centimeters.
The h intercept is when time is 0, so if you set t = 0 you obtain
h(0) = 490(0)(0) + 74(0) = 12
So the h intercept is 12 cm.
That is, the log was 12 cm off of the ground.
The t intercept is when the height is 0 so set h=0 and solve (factor) for t
0 = -490t^2 + 75t + 12
THe domain and range must be restricted to the real values of the life scenerio. Thus time must start at 0, so the domain starts at 0.
The range must only go as high as the vertex, but as low as the ground. (unless the frog could jump into a hole, thus below ground, thus below the x-axis)