Problem 61 - Calculus: One and Several Variables

Question

Determine the range of \(y=x^{2}-4 x-5\) (a) by writing \(y\) in the form \((x-a)^{2}+b\). (b) by first solving the equation for \(x\).

Step-by-Step Solution

Answer

(a) The range of \(y=x^{2}-4x-5\) when rewritten as \(y=(x-2)^2-9\) is \([-9, \infty)\). \n(b) Solving the equation for \(x\) confirms that the range of \(y=x^{2}-4x-5\) is all real numbers: \((-\infty, \infty)\)

Step 1: Rewrite in standard form

To rewrite \(y=x^{2}-4x-5\) in the form \((x-a)^{2}+b\), first identify the perfect square part of the equation. In this case, it's \(x^{2}-4x\). Half the coefficient of -4 is -2, so the perfect square is \((x-2)^2\), and when expanding, the constant is 4. Subtract this added 4 from the original -5 to get -9. So the equation in standard form is \(y=(x-2)^2-9\).

Step 2: Determine the range from standard form

In the form \((x-a)^2+b\), the range of y is all values \(\geq b\), provided the leading coefficient is positive. In this case, it is so the range is \([-9, \infty)\]

Step 3: Solve the quadratic equation

To solve the equation \(y=x^{2}-4x-5\) for x, first set \(y=0\) so you have \(x^{2}-4x-5=0\). Apply the quadratic formula \(x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}\) where \(a=1, b=-4, c=-5\). This will give you two solutions for x.

Step 4: Use solutions to determine range

The roots of the quadratic equation give the values of x for which y=0. The range of y will be all real numbers when considering the nature of quadratic equations as they extend indefinitely in either direction as \(x\rightarrow±\infty\)

Key Concepts

Standard Form

Understanding the standard form of a quadratic equation is essential to solving many algebraic problems. The standard form of a quadratic equation is expressed as \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). This form is valuable because it provides a straightforward way to identify the parabola's direction and calculate the vertex.

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

Rewriting a quadratic in the vertex form, \(y = (x-a)^2 + b\), as done in our example, simplifies finding the vertex. Here, the vertex is \((a, b)\), aiding in quickly determining key properties like the minimum or maximum value of the function.

Completing the Square

Completing the square is a technique used to rewrite quadratic functions in vertex form \((x-a)^2 + b\). It involves changing the equation to make one part a perfect square trinomial.
To complete the square for the equation \(y = x^2 - 4x - 5\):

• Focus on the terms \(x^2 - 4x\).
• Half the coefficient of \(x\) and square it, i.e., \((-4/2)^2 = 4\).
• Add and subtract this square inside the equation to keep the equation balanced: \(y = (x^2 - 4x + 4) - 4 - 5\).
• This transforms into \(y = (x-2)^2 - 9\).

This method helps easily identify the vertex form, clarifying the parabola's shape and helping determine the range more straightforwardly.

Quadratic Formula

The quadratic formula is a reliable method to find solutions (roots) for any quadratic equation. Expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), it requires substituting the values of \(a\), \(b\), and \(c\) from the general form \(ax^2 + bx + c = 0\).

In our example, for \(x^2 - 4x - 5 = 0\):

• \(a = 1\), \(b = -4\), and \(c = -5\)
• Discriminant = \((-4)^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36\)
• \(x = \frac{-(-4) \pm \sqrt{36}}{2 \cdot 1} = \frac{4 \pm 6}{2}\)
• Solutions are \(x = 5\) and \(x = -1\)

These roots determine the \(x\)-intercepts of the parabola, showing where the graph crosses the x-axis. It's an essential step in analyzing quadratic functions thoroughly.

Range of a Quadratic Function

The range of a quadratic function depends on its minimum or maximum value, aligned with its vertex when in vertex form. Quadratics create parabolic graphs that either open upwards or downwards, dictating whether the range expands upwards or downwards indefinitely.

For a function like \(y = (x-2)^2 - 9\), the range is determined from \((x-a)^2\), which always yields non-negative results. Since the parabola opens upwards, it means

• The lowest value of \((x-2)^2\) is 0, achieved when \(x = 2\).
• Thus, \(y = -9\) at the vertex, giving the range as \([-9, \infty)\).

This interpretation helps visualize and grasp how a quadratic function behaves across different \(x\) values, defining possible \(y\) values.