Question

Sketch the graph of the function. Label the vertex. y=-2 x^{2}-3 x+2

Short answer

The sketch of the function \(y=-2 x^{2}-3 x+2\) is a parabola opening downward, with its vertex at \((3/4, -2.125)\) and Y-intercept at \((0, 2)\).

Step-by-step solution

Find the vertex

The vertex of a quadratic function in the form \(y = a x^{2} + b x + c\) is given by the point \((-b/(2a), f(-b/(2a))\). Substituting the values \(a=-2\) and \(b=-3\), the x-coordinate of the vertex is given by \(-(-3)/(2*(-2)) = 3/4\). Substituting \(x=3/4\) into the function, the y-coordinate of the vertex is \(-2*(3/4)^{2} - 3*(3/4) + 2 = -2.125\). Hence, the vertex of the function is \((3/4, -2.125)\).

Find the Y-intercept

The Y-intercept of the function is the value of the function at \(x = 0\). Substituting \(x=0\) into the function gives \(y = -2*(0)^{2} - 3*0 + 2 = 2\). Hence, the Y-intercept of the function is at the point \((0, 2)\).

Sketch the graph

Now sketch the graph. Start by plotting the vertex \((3/4, -2.125)\) and the Y-intercept \((0,2)\). Since the coefficient of \(x^{2}\) is negative, the graph opens downward. Draw a graph that starts at the Y-intercept, reaches a maximum at the vertex, and then continues downward. Label the vertex.