Question

Sketch the graph of the function. Label the vertex. y=4 x^{2}-\frac{1}{4} x+4

Short answer

The graph of the function is an upward opening parabola, with a vertex at approximately (1/32, 7.9921) and a y-intercept at (0, 4). The vertex is labelled on the graph.

Step-by-step solution

Determine the vertex of the parabola

The vertex of a parabolic function \(y = ax^{2}+ bx + c\), is given by \((-b/2a, f(-b/2a))\). For this equation, \(a = 4\), and \(b = -1/4\). Therefore, the x-coordinate of the vertex is \(-(-1/4)/2*4 = 1/32\). Plugging this back into the function gives the y-coordinate as \(4(1/32)^2 - 1/4*1/32 + 4 = 4 - 1/128 + 4 = 7.9921\). Therefore, the vertex is approximately \((1/32, 7.9921)\).

Determine the direction of the parabola

The direction of the parabola is given by the sign of the coefficient a. Here, a = 4, which is positive. Hence, the parabola opens upwards.

Determine the y-intercept

The y-intercept is determined by setting x = 0 in the function. So, y = 4*0^2 - 1/4*0 +4 = 4.

Sketch the graph

Now, with the vertex, direction of parabola and y-intercept known, sketch the graph. Start by placing the vertex at (1/32, 7.9921) on the coordinate plane. Since the parabola is opening upwards, draw the parabola above the vertex, passing through the y-intercept (0, 4). Label the vertex.

Key concepts

Vertex of a Parabola

Understanding the vertex of a parabola is crucial when graphing quadratic functions. The vertex is the highest or lowest point on the graph of a parabola and serves as a pivotal element for its shape and position.

The formula to determine the vertex for the quadratic function in the form of \( y = ax^2 + bx + c \) is derived from completing the square or using calculus. However, a simplified method involves using the coordinates \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \). Here, \(a\) and \(b\) are coefficients from the quadratic equation.

In our exercise, the quadratic function is \( y = 4x^2 - \frac{1}{4}x + 4 \). By applying the vertex formula, we find that the x-coordinate is \( -\frac{-\frac{1}{4}}{2\times 4} = \frac{1}{32} \). To obtain the y-coordinate, we substitute \(x\) with \( \frac{1}{32} \) in the function, leading to a vertex at approximately \((\frac{1}{32}, 7.9921)\). This is where the parabola turns and thus is pivotal for the graphing process.

Direction of a Parabola

The direction that a parabola opens is determined by the sign of the coefficient \(a\) in the quadratic function \(y = ax^2 + bx + c\).

If \(a\) is positive, the parabola opens upward, resembling a cup that holds water. On the other hand, if \(a\) is negative, the parabola opens downward, resembling an umbrella. This is a quick way to determine the general shape of the parabola even before plotting any points.

In the given exercise, our equation has \(a = 4\), which is a positive number. Consequently, we can immediately conclude that the parabola will open upwards. This information assists us in visualizing the graph, forecasting that it will rise from the vertex, and allowing us to anticipate the nature of the function's maximum or minimum point.

Y-Intercept

The y-intercept of a quadratic function is the point where the graph intersects the y-axis. This happens when the value of \(x\) is zero.

To find the y-intercept algebraically, we simply replace \(x\) with zero in the quadratic equation. For the function \(y = 4x^2 - \frac{1}{4}x + 4\), setting \(x = 0\) yields \(y = 4\), revealing that the y-intercept is at point \((0, 4)\).

Knowing the y-intercept provides an additional point to plot on the graph and assists in constructing the overall shape of the parabola. Moreover, the y-intercept is a fundamental aspect of the graph as it also represents the initial value of the function when no other variables are influencing the outcome.