Question

Sketch the graph of the function. Label the vertex. y=\frac{1}{2} x^{2}+2 x-1

Short answer

The vertex of the function \(y = \frac{1}{2}x^2 + 2x - 1\) is at point (-2,-1). Graphed, this function is a parabola that opens upwards, with the vertex being the lowest point, and the y-intercept at \(y = -1\).

Step-by-step solution

Identify the quadratic function

The given function \(y = \frac{1}{2}x^2 + 2x - 1\) is a quadratic function, as it is of the form \(y = ax^2 + bx + c\). The coefficients are \(a = \frac{1}{2}\), \(b = 2\), and \(c = -1\). Quadratic functions form a U-shaped curve or a parabola when graphed.

Calculate the Vertex

The vertex of a quadratic function in standard form \(y = ax^2 + bx + c\) is given by \((-b/2a, f(-b/2a))\). For the given function, compute the x-coordinate of the vertex as \(-b/2a = -2/(2* \frac{1}{2}) = -2\). Substitute \(-2\) into the equation to obtain the y-coordinate: \(y = \frac{1}{2}*(-2)^2 + 2*(-2) - 1 = -1\). So, the vertex is \((-2, -1)\).

Graph the function

Start by drawing an x and y axis. Plot the vertex point that was found in the previous step. Because the coefficient of \(x^2\) is positive, the parabola opens upwards. Sketch the graph symmetrically around the vertex point. The y-intercept is found by substituting \(x = 0\) in the equation, which gives us \(y = -1\). A rough sketch of the parabola can be completed now.