use any method to locate the zeros of y = -3x^2 + 12x + 15
To locate the zeros of the given equation, y = -3x^2 + 12x + 15, we need to find the values of x for which y becomes zero.
1. Factoring:
First, let's try factoring the equation. However, keep in mind that in this particular case, factoring may not be possible because the quadratic coefficient (-3) and the constant term (15) do not have common factors.
y = -3x^2 + 12x + 15
= -3(x^2 - 4x - 5)
To continue factoring, we will look for two numbers whose product is -5 and whose sum is -4. The numbers -5 and 1 satisfy these conditions. So, the equation can be factored as:
y = -3(x - 5)(x + 1)
To find the zeros, we set y = 0:
0 = -3(x - 5)(x + 1)
Setting each factor to zero, we get:
x - 5 = 0 --> x = 5
x + 1 = 0 --> x = -1
Hence, the zeros of the equation are x = 5 and x = -1.
2. Quadratic Formula:
The quadratic formula can also be used to find the zeros of the equation in case factoring is not possible.
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, y = -3x^2 + 12x + 15, we have a = -3, b = 12, and c = 15. Substituting these values into the quadratic formula, we get:
x = (-(12) ± √((12)^2 - 4(-3)(15))) / (2(-3))
= (-12 ± √(144 + 180)) / (-6)
= (-12 ± √324) / (-6)
= (-12 ± 18) / (-6)
This gives us two possible solutions:
x = (-12 + 18) / (-6) = 1
x = (-12 - 18) / (-6) = 5
Therefore, the zeros of the equation are x = 1 and x = 5, which matches our previous method of factoring.