The temperature y (in degrees Fahrenheit) after t months can be modeled by the function y=-3t^2+18t+53 where 1<t<12.
A. Write the function in vertex form y=__
B. Find the maximum temperature during the year.
see your previous posts of the same problem.
note that
y = -3(t^2-6t) + 53
now complete the square to get the vertex form
To write the function in vertex form, we need to complete the square. The vertex form of a quadratic function is given by y = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.
A. Let's rewrite the given function in vertex form:
y = -3t^2 + 18t + 53
First, let's factor out the common factor of -3:
y = -3(t^2 - 6t) + 53
Now, we want to complete the square inside the parentheses. To do this, we need to take half of the coefficient of t (-6 in this case), square it, and add/subtract it inside the parentheses.
Step 1: Take half of the coefficient of t: -6/2 = -3
Step 2: Square the result: (-3)^2 = 9
Now, add and subtract this value inside the parentheses:
y = -3(t^2 - 6t + 9 - 9) + 53
Next, we group the terms inside the parentheses:
y = -3((t - 3)^2 - 9) + 53
Now, expand the parentheses:
y = -3(t - 3)^2 + 27 - 9 + 53
Simplify:
y = -3(t - 3)^2 + 71
So, the function in vertex form is y = -3(t - 3)^2 + 71.
B. The parabolic function is in the form y = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola. In this case, the vertex form is y = -3(t - 3)^2 + 71.
The vertex of the parabola occurs at the value t = 3, which is the h-coordinate. To find the maximum temperature, we need to find the k-coordinate.
In the given function, the coefficient of the squared term is -3, which is negative.
Since a is negative, the parabola opens downward, and the vertex represents the maximum point.
Therefore, the maximum temperature occurs when t = 3. Plugging this value into the function:
y = -3(3 - 3)^2 + 71
y = -3(0)^2 + 71
y = -3(0) + 71
y = 71
The maximum temperature during the year is 71 degrees Fahrenheit.