Given the quadratic equation y = -x^2 + 2x + 3.

a) Will the graph turn upwards or downwards?
b)What are the coordinates of x and y intercepts?
c)What is the coordinate of the turning point?
d)Use your answer in parts (a,b,c,and d) to sketch the graph.

a) the -x^2 tells you

b) y=(3-x)(x+1)
c) vertex at x = 1

a) To determine whether the graph of the quadratic equation will turn upwards or downwards, we need to look at the coefficient of the x^2 term. In this equation, the coefficient is -1 for the x^2 term. Since the coefficient is negative, the graph will open downwards.

b) To find the x-intercepts, we need to set y to zero and solve for x. In this equation, when y = 0, we have -x^2 + 2x + 3 = 0. We can solve this equation by factoring, completing the square, or using the quadratic formula.

Factoring: -x^2 + 2x + 3 = 0
=> (x - 3)(x + 1) = 0
=> x - 3 = 0 or x + 1 = 0
=> x = 3 or x = -1

The x-intercepts are (3, 0) and (-1, 0).

To find the y-intercept, we need to set x to zero and solve for y. When x = 0, we have y = -0^2 + 2(0) + 3 = 3. Therefore, the y-intercept is (0, 3).

c) The turning point of a quadratic equation is also known as the vertex. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. In this equation, a = -1 and b = 2. Therefore, x = -2/(2*-1) = 1.

To find the y-coordinate of the vertex, we substitute x = 1 into the equation y = -x^2 + 2x + 3: y = -(1)^2 + 2(1) + 3 = 4.

The coordinate of the turning point is (1, 4).

d) To sketch the graph, we can plot the x and y-intercepts, as well as the turning point. The graph will open downwards, with the vertex at (1, 4). The x-intercepts are (3, 0) and (-1, 0), and the y-intercept is (0, 3).