Problem 22 - 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. $$ \begin{aligned} &y=-x^{2}+4 x-2\\\ &B(3,-2) \end{aligned} $$

Step-by-Step Solution

Answer

The graph of the given function is a parabola opening downwards with its vertex at the point (2,2). The given point B (3,-2) is plotted outside this parabola.

Step 1: Graph the Parabola

Start by finding the vertex of the parabola. For a parabola of the form \(y=ax^{2}+bx+c\), the x-coordinate of the vertex is given by \(-b/2a\). Therefore, for this function, the x-coordinate of the vertex is \(-4/(-2) = 2\). Substitute this value of x into the function to find the y-coordinate of the vertex: \(y=-2^{2}+4*2-2=2\). So, the vertex of this parabola is at (2,2).\nThen, use plotting techniques to graph the function. Given that it's downside opening, mark the vertex and draw the parabola accordingly.

Step 2: Plot Point B

Plot the point B (3,-2) on the same graph.

Step 3: Determine the Location of the Point

To determine if point B is inside or outside the parabola, substitute the x-coordinate of point B into the equation and compare it to the y-coordinate. If y using B's x-coordinate is below the actual y-coordinate of B, it is outside. If above, it is inside: \(y=-3^{2}+4*3-2 = -9+12-2=1\). The y-coordinate using B's x is 1, which is above the actual y-coordinate of B (-2). Therefore, B lies outside the parabola.

Key Concepts

Parabola Vertex

The vertex of a parabola is a crucial point that helps to define its shape and orientation. Think of it as the "turning point" where the parabola switches directions. For a quadratic equation of the form \(y = ax^2 + bx + c\), the x-coordinate of the vertex is determined using the formula \(-b/2a\). In our specific example, we have the quadratic equation \(y = -x^2 + 4x - 2\). Here, \(a = -1\), \(b = 4\), and \(c = -2\). By applying the vertex formula, we find that the x-coordinate is \(-4/(-2) = 2\).
Substitute \(x = 2\) back into the quadratic equation to find the y-coordinate:

• Calculate \(y = -2^2 + 4 \times 2 - 2\).
• Perform the operations step by step: \(-4 + 8 - 2 = 2\).

Thus, the vertex is at \((2, 2)\). Understanding the vertex provides insight into where the parabola reaches its highest or lowest point, depending on the direction it opens. In this case, since the coefficient \(a\) is negative, the parabola opens downward.

Quadratic Equation

A quadratic equation is an equation that can be put in the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). This type of equation describes a parabola when graphed. The value of \(a\) determines the direction of the parabola:

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

Moreover, the coefficients \(b\) and \(c\) influence the position of the parabola on the graph. One critical aspect of quadratic equations is that they always form a parabola, a symmetrical shape. This symmetry is about the axis (the vertical line that passes through the vertex), which divides the parabola into two mirror-image halves. Recognizing the form of the quadratic equation is vital for sketching the graph accurately and analyzing its behavior.

Analyzing Graphs

Analyzing graphs involves interpreting the visual presentation of functions, such as parabolas. In this scenario, plotting the quadratic equation \(y = -x^2 + 4x - 2\) visually represents how the function behaves over different values of \(x\). Start by identifying the vertex and plotting key points, such as the vertex \((2, 2)\) and other points derived from the function.
Next, examine the direction the parabola opens. Since \(a\) is negative in this equation, the parabola opens downwards. This means it will reach a maximum point at the vertex before arching downwards.
For the point \(B(3, -2)\), assess its position by substituting its x-coordinate into the parabola's equation.

• Calculate \(y\) at \(x = 3: y = -3^2 + 4 \times 3 - 2 = 1\).

Compare the calculated \(y = 1\) with \(B's\) actual y-coordinate, \(-2\). Since 1 is greater than \(-2\), point \(B\) lies outside of the parabola. By following these steps, you can analyze and understand the position and orientation of any parabola on a graph.