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PLEASE HELP! WILL MARK BRAINLIEST!!
Describe the graph of the function f(x) = 3-x2+3x-2. Include the y-intercept, x-intercepts, and the shape of the graph.  Show all of your work for finding x and y-intercepts.

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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.

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