Problem 78 - Beginning and Intermediate Algebra

Question

Find the indicated values for the following polynomial functions. \(h(t)=t^{2}-6 t-16 .\) Find \(t\) so that $h(t)=0$$

Step-by-Step Solution

Answer

The values of \(t\) for which \(h(t) = 0\) are \(t = 8\) and \(t = -2\).

Step 1: Determine if the quadratic can be factored

First, we will try to factor the given quadratic function \(h(t) = t^2 - 6t - 16\). To do this, we must find two numbers that multiply to the constant term (-16) and add up to the middle term (-6). We'll soon realize that there are no two such numbers. Therefore, the quadratic is not factorable, so we'll use the quadratic formula to solve for \(t\) instead.

Step 2: Apply the quadratic formula

The quadratic formula is given by \(t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\), where the quadratic function takes the standard form of \(ax^2 + bx + c\). For our function, \(a=1\), \(b=-6\), and \(c=-16\). Substitute these values into the quadratic formula: \(t = \frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-16)}}{2(1)}\)

Step 3: Simplify the expression

Now, let's simplify the expression: \(t = \frac{6\pm\sqrt{36+64}}{2}\) \(t = \frac{6\pm\sqrt{100}}{2}\) Since the square root of 100 is 10, our expression becomes: \(t = \frac{6\pm10}{2}\)

Step 4: Calculate the values of \(t\)

Now, to find the values of \(t\) that make \(h(t) = 0\), we consider the two possibilities: 1) \(t = \frac{6 + 10}{2}\) \(t = \frac{16}{2}\) \(t = 8\) 2) \(t = \frac{6 - 10}{2}\) \(t = \frac{-4}{2}\) \(t = -2\) Therefore, the values of \(t\) that satisfy the equation \(h(t) = 0\) are \(t = 8\) and \(t = -2\).

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to express a quadratic equation, like \( ax^2 + bx + c \), as a product of two binomial expressions. The idea is to find two numbers that multiply to the constant term \( c \) and add up to the coefficient of the linear term \( b \).

For the quadratic \( h(t) = t^2 - 6t - 16 \), you would be looking for two numbers that multiply to \(-16\) and add to \(-6\). Sometimes, factoring can be straightforward.

• The numbers you seek must have a product equal to the last term (here \(-16\)).
• They should also sum up to the middle term coefficient (here \(-6\)).

In our example, no such numbers exist that satisfy both conditions. When this happens, it indicates that the quadratic cannot be factored into rational numbers.

Therefore, we must turn to other methods, like the quadratic formula, to solve the equation.

Quadratic Formula

The quadratic formula provides a way to solve quadratic equations that cannot be easily factored. It is expressed as: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a \), \( b \), and \( c \) are coefficients from the standard form \( ax^2 + bx + c \).

Using the quadratic formula involves these key steps:

• Identify the coefficients \( a \), \( b \), and \( c \) from the equation. For \( h(t) = t^2 - 6t - 16 \), we have \( a = 1 \), \( b = -6 \), and \( c = -16 \).
• Substitute these values into the quadratic formula.
• Simplify under the square root (called the discriminant), then simplify the entire expression further to find the solutions for \( t \).

In the example, after substituting and simplifying, the solutions found were \( t = 8 \) and \( t = -2 \). This method guarantees we can solve any quadratic equation, even if it's not easily factorable.

Solving Polynomial Equations

Solving polynomial equations involves finding the values of the variable that make the equation equal to zero. For quadratics, this typically boils down to finding the roots of the equation.

When faced with a polynomial equation like \( h(t) = t^2 - 6t - 16 \), you have several strategies:

• Start with factoring, if possible. This can quickly yield the roots.
• If factoring doesn't work (as in the case of our equation), apply the quadratic formula.
• Use graphical methods or algebraic rearrangements, which can also help, especially for higher degree polynomials.

By using these strategies, you can reliably find the zeros of polynomial equations, which are crucial in various mathematical applications.

Zero of a Function

The zero of a function is a value that makes the function equal to zero. For a quadratic function like \( h(t) = t^2 - 6t - 16 \), the zeros are the values of \( t \) that satisfy \( h(t) = 0 \).

Finding zeros is significant because they represent points where the graph of the function crosses or touches the t-axis. Understanding this gives insight into the function's behavior.

• Zeros for a quadratic equation can be found by either factoring or using the quadratic formula.
• In our example, \( t = 8 \) and \( t = -2 \) are the zeros. They are the solutions to the equation \( t^2 - 6t - 16 = 0 \).
• Zeros help determine the intervals where a function is positive or negative.

Thus, finding the zero of a function provides valuable information about that function's characteristics and behavior across its domain.