Question

Sketch the graph of the function. $$y=2 x^{2}-3 x+4$$

Short answer

The graph of the function \(y=2 x^{2}-3 x+4\) is a parabola that opens upwards, with its vertex at the point (0.75, 3.125) and an axis of symmetry at \(x = 0.75\).

Step-by-step solution

Find the Vertex

To find the vertex of the parabola, use the formula \(x=-\frac{b}{2a}\), where \(a\) is the coefficient of the \(x^2\) term and \(b\) is the coefficient of the \(x\) term. Substituting \(a = 2\) and \(b = -3\) into the formula, we get \(x = -\frac{-3}{2*2} = 0.75\). To find the y-coordinate of the vertex, substitute \(x = 0.75\) into the given equation: \(y = 2(0.75)^{2} - 3(0.75) + 4 = 3.125\). So, the vertex of the parabola is (0.75, 3.125).

Determine the Direction of the Parabola

The sign of \(a\) determines the direction in which the parabola opens. If \(a\) is positive, the parabola opens upwards; if \(a\) is negative, it opens downwards. In this case, as \(a = 2\), which is positive, our parabola opens upwards.

Find the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. As we've found that the x-coordinate of the vertex is 0.75, the equation for the axis of symmetry is \(x = 0.75\).

Sketch the Graph

Plot the vertex, draw the axis of symmetry, and sketch a rough figure of the graph that opens upwards with the vertex as the lowest point since the parabola opens upwards, respecting the symmetry over the axis of symmetry.