To locate the zeros of the quadratic function y = -3x^2 + 12x + 15, we can set y equal to zero and solve for x:
0 = -3x^2 + 12x + 15
To solve this quadratic equation, we can factor it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
From the equation y = -3x^2 + 12x + 15, we can see that a = -3, b = 12, and c = 15. Plugging these values into the quadratic formula, we get:
x = (-12 ± √(12^2 - 4(-3)(15))) / (2(-3))
x = (-12 ± √(144 + 180)) / (-6)
x = (-12 ± √324) / (-6)
x = (-12 ± 18) / (-6)
Simplifying further:
x = (-12 + 18) / (-6) or x = (-12 - 18) / (-6)
x = 6 / (-6) or x = -30 / (-6)
x = -1 or x = 5
Therefore, the zeros of the function y = -3x^2 + 12x + 15 are x = -1 and x = 5.
The correct answer is:
B. (-1, 0) (5, 0)
Use any method to locate the zeros of y = -3x^2 + 12x + 15.
A. (1, 0) (-5, 0)
B. (-1, 0) (5, 0)
C. There are no zeros.
D. (0, 15)
1 answer