Problem 24 - Algebra 1: Concepts and Skills

Question

Sketch the graph of the function. Plot the given point and determine whether the point lies inside or outside the parabola. $$ \begin{aligned} &y=4 x^{2}-x+1\\\ &D(-2,5) \end{aligned} $$

Step-by-Step Solution

Answer

Thus, the graph of the function is a parabola with vertex at (0.125, 1.03125) that opens upwards. The point (-2, 5) lies outside this parabola.

Step 1: Identify Elements of the Quadratic Function

The quadratic function is \(y = 4x^2 - x + 1\). So we know that \(a = 4\), \(b = -1\), and \(c = 1\).

Step 2: Find the Vertex of the Parabola

The vertex of the parabola \((h, k)\) is found using the formulas \(h = -\frac{b}{2a}\) and \(k = c - \frac{b^2}{4a}\). Substituting the given values of \(a\), \(b\), and \(c\) into these formulas, we find that the vertex \((h, k)\) of the parabola is \((h, k) = (0.125, 1.03125)\).

Step 3: Draw the Graph of the Parabola

Draw the graph of the function \(y = 4x^2 - x + 1\). Plot the vertex of the parabola and sketch the parabola. Remember that the parabola opens upward since \(a > 0\).

Step 4: Determine the Position of the Given Point

The given point is \((-2, 5)\). To check if this point lies on, inside, or outside the parabola, we substitute these 'x' and 'y' values into the equation \(y = 4x^2 - x + 1\). We find that \(5 < 4*(-2)^2 - (-2) + 1 = 21\). Hence, the point \((-2, 5)\) lies outside the parabola.

Key Concepts

Parabola Sketching

Sketching a parabola, the U-shaped curve represented by a quadratic function, involves identifying key features such as the vertex, axis of symmetry, and the direction in which the parabola opens.

To begin with, it's crucial to understand the general form of a quadratic equation, which is expressed as \(y = ax^2 + bx + c\). In our exercise, the quadratic function given is \(y = 4x^2 - x + 1\). Here, \(a = 4\), indicating that the parabola opens upwards because \(a > 0\). If \(a\) were less than zero, it would open downwards.

Knowing this, you start by finding the vertex, which serves as the highest or lowest point on the graph and is the point where the parabola changes direction. Once you've located the vertex, you then draw the axis of symmetry, a vertical line that passes through the vertex, dividing the parabola into two mirror images.

After plotting the vertex and drawing the axis of symmetry, you can sketch the parabola by plotting additional points around the vertex and ensuring that the curve is symmetrical. If more precision is required, finding other points by creating a table of values for \(x\) and \(y\) can help in sketching a more accurate graph.

Quadratic Function Vertex

The vertex of a quadratic function is an essential element since it represents the turning point of the parabola and is located at the axis of symmetry. To find the vertex \(h, k\), you can use the formulas \(h = -\frac{b}{2a}\) and \(k = c - \frac{b^2}{4a}\), derived from the quadratic function \(y = ax^2 + bx + c\).

In our example, \(a = 4\), \(b = -1\), and \(c = 1\), thus the calculations would be as follows: \(h = -\frac{-1}{2 \cdot 4} = 0.125\) and \(k = 1 - \frac{(-1)^2}{4 \cdot 4} = 1.03125\). This gives us the vertex \(0.125, 1.03125\), which is slightly to the right of the origin and above it, since both \(h\) and \(k\) are positive.

Identifying the vertex accurately is crucial as it aids in sketching a precise graph and provides valuable information about the function's maximum or minimum value, which in the context of real-life problems can represent critical data such as profit maximization or cost minimization in a business setting.

Position of Points Relative to Parabola

Determining the position of a point relative to a parabola involves assessing whether the point lies inside, outside, or directly on the parabolic curve. This can be done by substituting the coordinates of the point into the quadratic equation and comparing the resulting value with the original y-coordinate of the point.

Consider the given point \(D(-2, 5)\). To ascertain its position, we plug \(x = -2\) into the quadratic function \(y = 4x^2 - x + 1\) to compute the corresponding \(y\) value. If this calculated value is equal to the y-coordinate of the point \(D\), then the point lies on the parabola. If the computed \(y\) is less than the point's y-coordinate, the point is above the parabola; if greater, it's below. For our point \(D\), we find that \(5 < 4(-2)^2 - (-2) + 1 = 21\), therefore the point lies outside and above the parabola.

Understanding the relative position of points is vital in various applications, such as determining the feasibility of solutions in optimization problems, analyzing projectile motion in physics, or in graphical methods of solving systems of inequalities.