Problem 25 - Precalculus : Building Concepts and Connections

Question

Factor to find the \(x\)-intercepts of the parabola described by the quadratic function. Also find the real zeros of the function. $$G(t)=2 t^{2}-t-3$$

Step-by-Step Solution

Answer

The x-intercepts or real zeros of the function \(G(t)=2t^{2}-t-3\) are \(t=\frac{-3}{2}\) and \(t=1\).

Step 1: Factorize the equation

To factorize the quadratic function \(G(t)=2t^{2}-t-3\), which is in the standard form \(ax^{2}+bx+c\), we first need to find two numbers that add up to -1 (the coefficient of t) and multiply to -6 (product of 2 and -3). Those two numbers are -2 and 3, therefore the factorized equation becomes \(2t^{2} – 2t + 3t -3 = 2t(t -1) +3(t -1) = (2t+3)(t-1)\).

Step 2: Determine the x-intercepts

The x-intercepts or real zeros of the function are determined by setting the equation to 0 and solving for t. So we set \((2t+3)(t-1) = 0\). From this equation we find the roots \(t=\frac{-3}{2}\) and \(t=1\).

Step 3: Checking the result

We can check the result by substituting \(t=\frac{-3}{2}\) and \(t=1\) back into the original equation to ensure it equals to 0.

Key Concepts

Factoring Quadratics

Factoring a quadratic equation involves rewriting it in a product form. Specifically, for a standard quadratic equation like the one given, the goal is to express it as a product of two simpler binomials. In the equation \[ G(t) = 2t^2 - t - 3 \]we look for two numbers, say \( p \) and \( q \), that multiply to the product of the coefficient of \( t^2 \) (or \( a \), which is 2 here) and the constant term, which is -3. Therefore, we need numbers that multiply to -6 and add up to -1, the coefficient of \( t \). These numbers are -2 and 3.
We then use these two numbers to split the middle term, \(-t\), into \[ 2t^2 - 2t + 3t - 3 \]
This allows us to factor by grouping: \[ 2t(t - 1) + 3(t - 1) \]which results in:\[ (2t + 3)(t - 1) \]
This is the factored form of the given quadratic equation, and makes solving for roots much simpler.

X-Intercepts

Once the quadratic function is factored, finding the x-intercepts is straightforward. Here, we're looking for the values of \( t \) (or \( x \), in general scenarios) that make the function equal to zero. These points are where the graph of the quadratic equation crosses the \( t \)-axis. Given the factored form:\[ (2t + 3)(t - 1) = 0 \]
To find the intercepts, set each factor equal to zero and solve for \( t \). This gives:

• \( 2t + 3 = 0 \) => \( t = \frac{-3}{2} \)
• \( t - 1 = 0 \) => \( t = 1 \)

Thus, the x-intercepts are \( t = \frac{-3}{2} \) and \( t = 1 \). These points are crucial since they provide information about where the quadratic function crosses or touches the \( t \)-axis.

Real Zeros

In the context of quadratic functions, "real zeros" refer to the solutions of the equation when set to zero. These solutions are the same as the x-intercepts. They represent the values of \( t \) for which \( G(t) = 0 \). Not all quadratics will have real zeros, as some may have complex results instead without any crossing of the x-axis, but our function does.
The factored form \((2t + 3)(t - 1) = 0 \) leads us to find the real zeros exactly as we find the x-intercepts. Solving gives us:

• \( t = \frac{-3}{2} \)
• \( t = 1 \)

These are the real zeros of the function \( G(t) \), indicating that these are the points where the function equals zero. This concept links directly to the function's roots and intercepts, providing a comprehensive understanding of its behavior.