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 previous post.
A. To write the function in vertex form, we need to complete the square. The vertex form of a quadratic function is given by the equation y = a(t - h)^2 + k, where (h, k) represents the vertex of the parabola.
First, let's rewrite the given function as y = -3t^2 + 18t + 53. To complete the square, we need to divide the coefficient in front of t by 2, square it, and add it to both sides of the equation:
y = -3(t^2 - 6t) + 53
y = -3(t^2 - 6t + 9) + 53 + 3(9)
y = -3(t - 3)^2 + 53 + 27
y = -3(t - 3)^2 + 80
So, in vertex form, the function is y = -3(t - 3)^2 + 80.
B. In the vertex form, the maximum temperature occurs at the vertex, which is the point (h, k). In this case, the vertex is (3, 80). Thus, the maximum temperature during the year is 80 degrees Fahrenheit.
To find the function in vertex form, we need to complete the square for the given quadratic equation.
The general form of a quadratic equation is y = ax^2 + bx + c.
In this case, a = -3, b = 18, and c = 53.
To complete the square, we need to find the value that completes the square for the quadratic term. This is found by taking half of the coefficient of the x term (b/2) and then squaring it.
For the given equation, b/2 = 18/2 = 9. Squaring this value gives us 9^2 = 81.
Next, we add this value to both sides of the equation to keep it balanced:
y - 81 = -3t^2 + 18t + 53 + 81
Simplifying further:
y - 81 = -3t^2 + 18t + 134
Next, we can rewrite the quadratic term in vertex form:
y - 81 = -3(t^2 - 6t) + 134
Now, we need to complete the square for the expression inside the parentheses. This involves adding and subtracting the square of half of the coefficient of the linear term (-6/2 = -3) squared (-3^2 = 9):
y - 81 = -3(t^2 - 6t + 9 - 9) + 134
Simplifying further:
y - 81 = -3((t - 3)^2 - 9) + 134
Expanding the equation:
y - 81 = -3(t^2 - 6t + 9) + 27 + 134
Combining like terms:
y - 81 = -3(t^2 - 6t + 9) + 161
Finally, we can factor out the -3 from the expression inside the parentheses:
y - 81 = -3(t - 3)^2 + 161
So, the function in vertex form is y = -3(t - 3)^2 + 161.
Now, to find the maximum temperature during the year, we just need to look at the vertex of the quadratic function. The vertex of a quadratic function in the form y = a(x - h)^2 + k is given by (h, k).
In this case, the vertex is (3, 161). The x-coordinate of the vertex represents the month, and the y-coordinate represents the temperature.
Therefore, the maximum temperature during the year is 161 degrees Fahrenheit.