Problem 66 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ y=4 x^{2}-32 x+63 $$

Step-by-Step Solution

Answer

The parabola's vertex is at (4, -1).

Step 1: Identify the Equation Type

The given equation is in the form of a quadratic equation: \( y = 4x^2 - 32x + 63 \). This indicates that the graph is a parabola. Since \( y \) is expressed in terms of \( x \), this is a vertical parabola.

Step 2: Complete the Square

To write the equation in standard form, complete the square. Start with the equation: \[ y = 4x^2 - 32x + 63 \] Factor out the 4 from the \( x^2 \) and \( x \) terms:\[ y = 4(x^2 - 8x) + 63 \]Next, complete the square for the expression in the parentheses. Take half of the coefficient of \( x \) (which is -8), square it, and add inside the parentheses:\[ x^2 - 8x \rightarrow (x - 4)^2 - 16 \]Thus, the equation becomes:\[ y = 4((x - 4)^2 - 16) + 63 \]

Step 3: Simplify the Equation

Now distribute the 4 and simplify:\[ y = 4(x - 4)^2 - 64 + 63 \]Combine like terms:\[ y = 4(x - 4)^2 - 1 \]The equation \( y = 4(x - 4)^2 - 1 \) is now in the standard form of a parabola, \( y = a(x-h)^2 + k \).

Step 4: Identify the Vertex

In the standard form \( y = a(x-h)^2 + k \), the vertex of the parabola is at \((h, k)\). From the equation:\[ y = 4(x - 4)^2 - 1 \]The vertex of the parabola is \((4, -1)\).

Step 5: Graph the Parabola

To graph the parabola, plot the vertex at \((4, -1)\). Since the leading coefficient \( a = 4 \) is positive, the parabola opens upwards. Use other points to further define the curve as needed.

Key Concepts

Standard Form

Typically, a quadratic equation can be manipulated into a specific kind of expression known as the standard form. The standard form of a parabola is given by the equation:\[ y = a(x-h)^2 + k \]Here:

• \(a\) is a non-zero constant that dictates the direction and width of the parabola.
• \(h\) and \(k\) are the coordinates of the vertex.

When you have an equation like \(y = 4x^2 - 32x + 63\), it's useful to express it in this standard form to easily read off the vertex coordinates, and understand the parabola's shape and orientation. By completing the square, you reorganize the quadratic into this tidy format. This step clarifies the vertex's location and the direction in which the parabola opens.

Vertex

The vertex of a parabola is a crucial point that indicates its highest or lowest depending on whether it opens upwards or downwards. If you manage to write a quadratic equation in the standard form \(y = a(x-h)^2 + k\), it's simple to pinpoint its vertex directly from the equation.

• The vertex is found at the coordinates \((h, k)\).

In our context with the final standard form \(y = 4(x - 4)^2 - 1\), the vertex is clearly at \((4, -1)\). Having the vertex helps in accurately sketching the graph of the parabola and identifying the axis of symmetry. More practically, it defines a crucial point from which the parabola extends in either an upward "U" shape or a downward arch depending on the sign of \(a\).

Quadratic Equation

Quadratic equations are polynomial equations of the second degree that take on the general form:\[ ax^2 + bx + c = 0 \]These equations are key in algebra and often used to model various real-world scenarios. In the case of the equation \(y = 4x^2 - 32x + 63\), we're dealing with a quadratic expressed with a function form that describes a parabola.

• The coefficient \(a\), when positive, tells us the parabola opens upwards.
• The interaction between \(a\), \(b\), and \(c\) can be creatively shifted into a vertex form for easier use in graphing.

Solving this involves methods like factoring, using the quadratic formula, or completing the square, all of which are foundational in finding roots or zeros of the equation, as well as turning it into a more visually interpretable form. Here, organizing it until it becomes \(y = 4(x - 4)^2 - 1\) bridges a gap to graphing it directly, using the vertex and its formula-supported insights.