What is the axis of symmetry for F(X)= (x+2) (x-8)?
it will be midway between the two roots.
That is at x = (-2 + 8)/2
What is the symmetry and degree of the function f(x)=2x^3-4x^2-3x?
To find the axis of symmetry for a quadratic function, you can use the formula x = -b / (2a), where a represents the coefficient of the x^2 term and b represents the coefficient of the x term.
In the given function, F(x) = (x + 2)(x - 8), the coefficient of the x^2 term is 1, and the coefficient of the x term is 0:
a = 1
b = 0
Now we can substitute these values into the formula to find the axis of symmetry:
x = -b / (2a)
= -0 / (2 * 1)
= 0 / 2
= 0
Therefore, the axis of symmetry for the function F(x) = (x + 2)(x - 8) is x = 0.
To find the axis of symmetry for a quadratic function, you need to use the formula x = -b/2a, where the quadratic equation is in the standard form of ax^2 + bx + c = 0. In this case, the quadratic function is f(x) = (x + 2)(x - 8).
First, let's rewrite the function in standard form:
f(x) = x^2 - 6x - 16
Comparing this to the standard form ax^2 + bx + c, we see that a = 1, b = -6, and c = -16.
Now we can find the axis of symmetry:
x = -b/2a
x = -( -6 ) / ( 2 * 1 )
x = 6 / 2
x = 3
Therefore, the axis of symmetry for the function f(x) = (x + 2)(x - 8) is x = 3.