Question

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=-x^{2}-2 x+3\)

Short answer

The graph of the function \( y = -x^{2} - 2x + 3 \) is a downward-opening parabola with its vertex at (-1, 4) and roots at (1, 0) and (-3, 0).

Step-by-step solution

Find the Vertex of the Quadratic Equation

The quadratic equation is in the form \( y = ax^2 + bx + c \). The vertex of the equation is given as \( h = -b/2a \). In this case, \( a = -1, b = -2 \), so the \( h = - (-2) / (2*(-1)) = -1 \). Substitute this value into the equation \( y = -(-1)^2 - 2*(-1) +3 \) to find \( k = -1 + 2 + 3 = 4 \). So the vertex of the equation is at point (-1, 4).

Find the Axis of Symmetry

The Axis of Symmetry of a quadratic function is the vertical line that passes through the vertex. Therefore, for our function, the axis of symmetry is x = -1.

Find the Roots of the Equation

The roots of the equation are found by setting \( y = 0 \) i.e. \( -x^{2} - 2x + 3 = 0 \). Solving for \( x \) we get \( x = -1 \pm \sqrt{(-1)^2 - 4*1*(-3)} / 2*1 \). Hence, the roots of the equation are \( x = -1 + \sqrt{4} \) and \( x = -1 - \sqrt{4} \), yielding \( x = 1 \) and \( x = -3 \).

Sketch the Graph

Start by plotting the vertex at (-1, 4). Next, draw a vertical line (a dashed line usually) through the vertex, this is the axis of symmetry (x = -1). Then plot the roots at (1, 0) and (-3, 0). The parabola opens downwards since the coefficient of \( x^2 \) is negative. Connect the points with a smooth curve to form the graph of the function.