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. \(y=2 x^{2}-4 x+1\)

Short answer

The graph of the function \(y=2x^{2}-4x+1\) is a parabola opening upwards. The vertex of the graph is at point (1,-1) and the axis of symmetry is \(x=1\).

Step-by-step solution

Analyze the parabolic function

A parabolic function of form \(y=ax^{2}+bx+c\) has a vertex \((h,k)\) determined by the formula \(h= -b/2a\). With \(a=2\) and \(b=-4\), calculate the value of \(h=-(-4)/2*2 = 1\).

Identify Vertex point

Now, using \(h=1\), get the \(y\)-coordinate of which is the value of \(k\), by substituting \(h\) into the function: \(k=2*(1)^{2}-4*1+1 = -1\). The vertex of the function is \((1,-1)\).

Identify Axis of symmetry

The axis of symmetry of a parabola equation \(y = ax^{2} + bx + c\) is the vertical line \(x = h\). So, for this function, the axis of symmetry is \(x=1\).

Sketch the graph

Plot the points, make sure to mark the vertex and axis of symmetry. Start at the vertex (1,-1), and because the coefficient \(a\) of \(x^{2}\) is positive, the parabola opens upwards. Draw a smooth, u-shaped curve.