Use any method to locate the zeros of y=−3x2+12x+15 .(1 point) Responses (1,0)(−5,0) left parenthesis 1 comma 0 right parenthesis left parenthesis negative 5 comma 0 right parenthesis There are no zeros. There are no zeros. (0,15) left parenthesis 0 comma 15 right parenthesis (−1,0)(5,0) left parenthesis negative 1 comma 0 right parenthesis left parenthesis 5 comma 0 right parenthesis
To locate the zeros of the quadratic equation y = -3x^2 + 12x + 15, we can set y equal to zero and solve for x.
0 = -3x^2 + 12x + 15
Dividing the equation by -3 to simplify, we get:
0 = x^2 - 4x - 5
Next, we can factor the equation:
0 = (x - 5)(x + 1)
Setting each factor equal to zero and solving for x, we find the zeros:
x - 5 = 0 ---> x = 5
x + 1 = 0 ---> x = -1
Therefore, the zeros of the equation are (5,0) and (-1,0).
To locate the zeros of the quadratic function y = -3x^2 + 12x + 15, we can use the quadratic formula. The quadratic formula states that for a quadratic equation in the form ax^2 + bx + c = 0, the zeros can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -3, b = 12, and c = 15. Substituting these values into the quadratic formula:
x = (-12 ± √(12^2 - 4(-3)(15))) / (2(-3))
Simplifying further:
x = (-12 ± √(144 + 180)) / (-6)
x = (-12 ± √324) / (-6)
x = (-12 ± 18) / (-6)
This gives us two possible solutions for x:
x = (-12 + 18) / (-6) = 6 / (-6) = -1
x = (-12 - 18) / (-6) = -30 / (-6) = 5
Therefore, the zeros of the quadratic function y = -3x^2 + 12x + 15 are (-1, 0) and (5, 0).
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.
To do this, we can use the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, -3x^2 + 12x + 15 = 0, we can identify 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))
Simplifying further:
x = (-12 ± √(144 + 180)) / -6
x = (-12 ± √324) / -6
x = (-12 ± 18) / -6
Now, we can calculate the two possible values for x:
x1 = (-12 + 18) / -6 = 6 / -6 = -1
x2 = (-12 - 18) / -6 = -30 / -6 = 5
Therefore, the zeros of the equation y = -3x^2 + 12x + 15 are (-1, 0) and (5, 0).