Problem 34 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=2 x^{2}-x+1$$

Step-by-Step Solution

Answer

Vertex of the function \(f(x) = 2x^2 - x + 1\) is \(\left(\frac{1}{4}, \frac{15}{16}\right)\). Y-intercept is \(1\) and the function has no real roots, so it has no x-intercepts.

Step 1: Find the vertex

To find the vertex of the given quadratic function, first identify a, b from the general quadratic format \(ax^{2} + bx + c\). Here, \(a = 2\) and \(b = -1\). Plug these values into the vertex formula \(h = -\frac{b}{2a}\) to find \(h\), then use \(k=f(h)\) to find \(k\). This way we get the coordinates for the vertex.

Step 2: Find the x-intercept

Set \(f(x) = 0\) and solve for x to find the x-intercept(s). That is, solve the equation \(2x^{2}-x+1=0\). If the equation has real roots, these are the x-intercepts. If it doesn't have real roots, then the parabola does not intersect the x-axis and thus has no x-intercept.

Step 3: Find the y-intercept

Set \(x = 0\) and solve for \(f(x)\) to find the y-intercept. The y-coordinate can be found by just evaluating the function at \(x = 0\). This gives the y-intercept.

Step 4: Sketch the graph

Using the vertex, the y-intercept and the possible x-intercepts obtained in the previous steps, sketch the graph of the quadratic function.

Key Concepts

Vertex of a Quadratic

The vertex of a quadratic function is the highest or lowest point on its graph, which is a parabola. It's a major feature of the graph because it indicates the function's maximum or minimum value.
To find the vertex of the quadratic function f(x) = 2x^2 - x + 1, we take the coefficients a = 2 and b = -1 and use the vertex formula h = -b/(2a). Calculating it, h = -(-1)/(2*2) = 1/4. The x-coordinate of the vertex is thus 1/4. To get the y-coordinate, we evaluate the function at h: k = f(1/4), resulting in k = 2*(1/4)^2 - (1/4) + 1, which simplifies to the exact value that represents the vertex's y-coordinate. The complete vertex of the parabola for this function is at (1/4, k), which is crucial when sketching the quadratic graph.

X-intercept

The x-intercept(s) of a graph are the points where it crosses the x-axis. For quadratics, these represent the solution to the equation when f(x) = 0.
To find the x-intercept(s) for f(x) = 2x^2 - x + 1, we solve 2x^2-x+1=0. This quadratic equation might have two solutions, one solution, or none, depending on the discriminant b^2 - 4ac.
In our example, the discriminant is (-1)^2 - 4*2*1 = 1 - 8 = -7, which is less than 0. So, this function has no real x-intercepts, as the parabola doesn't cross the x-axis.

Y-intercept

The y-intercept is the point where a graph meets the y-axis, and for any function, it happens when x = 0. For the function f(x) = 2x^2 - x + 1, we find the y-intercept by evaluating f(0).
Substituting x with 0, we get f(0) = 2*(0)^2 - (0) + 1 = 1. This means that the y-intercept of the graph is at the point (0, 1). When sketching the graph, the y-intercept serves as an anchor point and helps us understand where the parabola passes with respect to the y-axis.

Quadratic Function Graph

A quadratic function graph takes the shape of a parabola which opens upwards if a > 0 and downwards if a < 0. As discussed, the vertex, x-intercept, and y-intercept inform the shape and position of this parabola on the coordinate plane.
For f(x) = 2x^2 - x + 1, because a = 2 is positive, the parabola opens upwards. Since we know the vertex and y-intercept, and that there are no real x-intercepts, we'd sketch an upward-opening parabola that has its vertex just above the x-axis, passing through the y-intercept at (0, 1).

Drawing the Graph

Plot the vertex and the y-intercept, then draw a smooth curve passing through these points, winging upwards perpetually on both sides, to represent the parabola of the given quadratic function.