Answer:
The y-intercept is (0,1). The x values at the horiz. intercepts are:
{ (-3+√5)/2, (-3-√5)/2 }. Graph is that of a parabola that opens down.
Step-by-step explanation:
It'd be helpful to rewrite f(x) = 3-x2+3x-2 with the powers of x in descending order and using " ^ " to denote exponentiation:
f(x) = -x^2 + 3x -2 + 3, or
f(x) = -x^2 + 3x + 1
y-intercept: let x = 0. Then f(0) = 1. The y-intercept is (0,1)
x-intercepts: let y = f(x) = 0 = -x^2 + 3x + 1. Let's use the quadratic formula with a = -1, b = 3 and c = 1 to determine the roots of this polynomial. The discriminant, b^2-4ac, is 9-4(1)(1), or 5.
Thus, there are two real, unequal roots. They are:
-3 plus or minus √5
x = ---------------------------------
2
or { (-3+√5)/2, (-3-√5)/2 }
These are your two horizontal intercepts, where the function is zero.
Because the coefficient -1 of the x^2 term is negative, the parabola representing this function opens down.
