Question 23687
Find the vertex:
The x-coordinate of the vertex of a parabola is given by: {{{x = -b/2a}}}
Your quadratic equation is:
{{{y = -x^2+2x+3}}} where: a = -1, b = 2, and c = 3

The x-coordinate of the vertex is:
{{{x = -2/2(-1)}}}
{{{x = 1}}}

The y-coordinate is found by substituting this value of x (1) into the quadratic equation and solving for y.
{{{y = -(1)^2 + 2(1) + 3}}}
{{{y = -1 + 2 + 3}}}
{{{y = 4}}}

The vertex is at: (1, 4)

The (x) intercepts can be found by solving the quadratic equation for the two root (where the parabola crosses the x-axis, if the roots are real).
{{{y = -x^2 + 2x + 3}}} Set y = 0 because this is where the parabola will cross the x-axis.
{{{-x^2 +2x + 3 = 0}}} Factor -1 from the left side.
{{{-(x^2 - 2x - 3) = 0}}} Factor the parentheses.
{{{-(x - 3)(x + 1) = 0}}} Clear the -1 by dividing both sides by -1
{{{(x - 3)(x + 1) = 0}}} Apply the zero product principle.
{{{x - 3 = 0}}} and/or {{{x + 1 = 0}}}

The roots (x-intercepts) are:
x = 3
x = -1

Here's the graph:

{{{graph(300,200,-5,5,-5,6,-x^2+2x+3)}}}