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.
y = -3(t^2-6t) + 53
= -3(t^2-6t+9) + 53 +3(9)
= -3(t-3)^2 + 71
To write the function in vertex form, we need to complete the square. The standard form of a quadratic equation is y = at^2 + bt + c, where a, b, and c are constants.
A. To complete the square, we need to find the value of the constant c. In this case, c = 53. To find the terms that complete the square, we take half of the coefficient of t and square it.
The coefficient of t is b = 18. Half of 18 is 9, and 9 squared is 81. So, we need to add 81 to the equation to complete the square. However, since we are adding 81, we need to subtract it as well to maintain the original equation balance.
The vertex form is y = a(t - h)^2 + k, where (h, k) represents the coordinates of the vertex.
Therefore, y = -3t^2 + 18t + 53 can be rewritten as y = -3(t^2 - 6t + 9) + 53 - 81. Simplifying further, we get y = -3(t - 3)^2 - 28.
So, the function in vertex form is y = -3(t - 3)^2 - 28.
B. In vertex form, the vertex represents the maximum or minimum point of the quadratic function. In this case, the vertex has the coordinates (h, k) = (3, -28).
Therefore, the maximum temperature occurs after 3 months, with a value of -28 degrees Fahrenheit.