Find a formula for a quadratic function whose graph has a y-intercept at y = 21 and only one zero at x = –28.
To find a formula for a quadratic function with the given information, we can start by using the general form of a quadratic equation:
f(x) = ax^2 + bx + c
We know that the y-intercept occurs at y = 21, which means when x = 0, f(x) = 21. Substituting these values into the equation, we get:
21 = a(0)^2 + b(0) + c
21 = c
So, we have found that c = 21. Now we can focus on finding the values of a and b.
Given that the quadratic function has only one zero at x = -28, we can determine the value of a using the fact that the quadratic has a double root at that point. A double root occurs when the discriminant of the quadratic equation is zero.
The discriminant is given by:
b^2 - 4ac
We substitute the values we know into the discriminant formula:
(-28)^2 - 4a(21) = 0
784 - 84a = 0
Rearranging the equation:
84a = 784
a = 784 / 84
a = 28 / 3
We have found that a = 28 / 3. Now we can find b by substituting the values of a and c into the quadratic equation:
f(x) = (28/3)x^2 + bx + 21
Let's use the fact that the quadratic has a single zero at x = -28. Since x = -28 is a zero, it means that when we substitute x = -28 into the equation, the result should be 0:
0 = (28/3)(-28)^2 + b(-28) + 21
Simplifying the equation:
0 = (784/3)(28) - 28b + 21
0 = 21952/3 - 28b + 21
To solve for b, we can isolate the term with b on one side of the equation:
28b = 21952/3 - 21
28b = (21952 - 63) / 3
28b = 21889 / 3
b = 21889 / (3 * 28)
b = 103.8928571
Now that we have the values of a, b, and c, we can write the formula for the quadratic function:
f(x) = (28/3)x^2 + (103.8928571)x + 21