find the axis of symmetry f(x)=-2x2-12x+8find the axis of symmetry f(x)=-2x2-12x+8
f(x)=2(x^2-6x+4)=2(x^2-6x+9-5)=
=2((x-3)^2-5)=2(x-3)^2-10
The vertical axis x=3
To find the axis of symmetry of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula x = -b / 2a.
In the given function, f(x) = -2x^2 - 12x + 8, a = -2 and b = -12.
Now, substitute these values into the formula x = -b / 2a:
x = -(-12) / (2 * -2)
x = 12 / -4
x = -3
Therefore, the axis of symmetry for the function f(x) = -2x^2 - 12x + 8 is x = -3.
To find the axis of symmetry of a quadratic equation in the form f(x) = ax^2 + bx + c, you can use the formula x = -b / (2a).
In this case, the equation is f(x) = -2x^2 - 12x + 8. Comparing it to the standard form, we have a = -2, b = -12, and c = 8.
Now, substitute these values into the formula x = -b / (2a):
x = -(-12) / (2 * -2)
= 12 / (-4)
= -3
Therefore, the axis of symmetry for the given quadratic equation is x = -3.