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.
A. To write the function in vertex form, we can complete the square. The vertex form of a quadratic equation is given by:
y = a(t - h)^2 + k
Let's start by completing the square for the given equation:
y = -3t^2 + 18t + 53
To complete the square, we need to factor the coefficient of t^2, which is -3. Divide the entire equation by -3 to make the coefficient of t^2 equal to 1:
y = (-3/(-3))t^2 + (18/(-3))t + (53/(-3))
y = t^2 - 6t - (53/3)
Now, we need to rearrange the equation by grouping the t terms:
y = (t^2 - 6t) - (53/3)
To complete the square, we take half of the coefficient of the t term, square it, and add it to both sides of the equation:
y = (t^2 - 6t + 9) - (53/3) - 9
y = (t - 3)^2 - 62/3
Therefore, the function in vertex form is:
y = (t - 3)^2 - 62/3
B. In the vertex form, the maximum value of y is equal to the value of k. Therefore, the maximum temperature during the year is -62/3 degrees Fahrenheit.
A. 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) is the vertex of the parabola.
The original function is y = -3t^2 + 18t + 53.
To complete the square, we'll focus on the t^2 and t terms.
Step 1: Divide the coefficient of the t term by 2, and then square the result.
(18/2)^2 = 81
Step 2: Add the result from step 1 to both sides of the equation.
y + 81 = -3t^2 + 18t + 53 + 81
y + 81 = -3t^2 + 18t + 134
Step 3: Rewrite the right side as a perfect square trinomial.
y + 81 = -3(t^2 - 6t + 9) + 134
Step 4: Factor the perfect square trinomial inside the parentheses.
y + 81 = -3(t - 3)^2 + 134
Step 5: Simplify the equation.
y = -3(t - 3)^2 + 134 - 81
y = -3(t - 3)^2 + 53
Therefore, the function in vertex form is y = -3(t - 3)^2 + 53.
B. In vertex form, the vertex of the parabola is given by (h, k). For this equation y = -3(t - 3)^2 + 53, the vertex is (3, 53).
Since the coefficient of the t^2 term is negative, the graph of the function is a downward-opening parabola.
Therefore, the maximum temperature during the year occurs at the vertex, which is (3, 53).
Hence, the maximum temperature during the year is 53 degrees Fahrenheit.
To write the function in vertex form, we need to complete the square by following these steps:
Step 1: Group the t terms together:
y = -3t^2 + 18t + 53
Step 2: Factor out the coefficient of t^2:
y = -3(t^2 - 6t) + 53
Step 3: Take half of the coefficient of t (which is -6), square it, and add it inside the parentheses. Add the same amount outside the parentheses to maintain equality:
y = -3(t^2 - 6t + 9 - 9) + 53
Step 4: Simplify the expression inside the parentheses by factoring and combining like terms:
y = -3((t - 3)^2 - 9) + 53
Step 5: Distribute the -3 to both terms inside the parentheses:
y = -3(t - 3)^2 + 27 + 53
Step 6: Simplify the constant terms:
y = -3(t - 3)^2 + 80
So, the function in vertex form is y = -3(t - 3)^2 + 80.
To find the maximum temperature during the year, we can directly read the y-coordinate of the vertex from the vertex form. In this case, the maximum temperature will occur at the vertex of the parabola, which is the point (h, k) = (3, 80).
Therefore, the maximum temperature during the year is 80 degrees Fahrenheit.
a. Y = a(x-h)^2 + k.
h = -b/2a = -18/-6 = 3,
k = -3*3^2 +18*3 + 53 = 80,
Y = -3(x-3)^2 + 80.
b. T max = k = 80o F.