Problem 30 - A Graphical Approach to College Algebra

Question

It is not apparent from the standard viewing window whether the graph of the quadratic function intersects the \(x\) -axis once, twice, or not at all. Experiment with various windows to find the number of \(x\) -intercepts. If there are \(x\) -intercepts, give their coordinates to the nearest hundredth. (GRAPH CANT COPY) $$y=x^{2}+6.95 x+12.07$$

Step-by-Step Solution

Answer

The graph intersects the x-axis at \((-3.40, 0)\) and \((-3.55, 0)\).

Step 1: Determine the Nature of the Roots

Calculate the discriminant using the formula \( b^2 - 4ac \) from the quadratic equation \( ax^2 + bx + c \). Here, \( a = 1 \), \( b = 6.95 \), and \( c = 12.07 \). The discriminant is calculated as:\[d = 6.95^2 - 4 imes 1 imes 12.07\]Calculate \(d\) to determine if the roots are real and how many roots exist.

Step 2: Evaluate the Discriminant

Compute the value of the discriminant from Step 1: \[d = 6.95^2 - 4 imes 1 imes 12.07 = 48.3025 - 48.28 = 0.0225\]Since the discriminant is greater than zero, there are two distinct real roots, indicating two \(x\)-intercepts.

Step 3: Find the Coordinates of the x-Intercepts

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots:\[x_1, x_2 = \frac{-6.95 \pm \sqrt{0.0225}}{2 imes 1}\]Calculate the roots to find the exact \(x\)-intercepts.

Step 4: Calculate the x-Intercepts

Compute the \(x\)-intercepts:\[x_1 = \frac{-6.95 + 0.15}{2} = \frac{-6.80}{2} = -3.40\]\[x_2 = \frac{-6.95 - 0.15}{2} = \frac{-7.10}{2} = -3.55\]Thus, the \(x\)-intercepts are at \((-3.40, 0)\) and \((-3.55, 0)\).

Key Concepts

Discriminant

The discriminant is a key concept when working with quadratic functions. It helps us understand the nature of the roots of a quadratic equation, which takes the form \( ax^2 + bx + c \). To find the discriminant, we use the formula \( b^2 - 4ac \).
This provides critical information about the solutions of the equation.

• If the discriminant is positive, there are two distinct real roots, meaning the quadratic function has two \(x\)-intercepts.
• If the discriminant is zero, there is exactly one real root, indicating that the graph of the quadratic forms a tangent to the \(x\)-axis at this point.
• If the discriminant is negative, there are no real roots. Consequently, the quadratic function does not intersect the \(x\)-axis.

In our equation, \( y = x^2 + 6.95x + 12.07 \), the discriminant is calculated as \( 6.95^2 - 4 \times 1 \times 12.07 = 0.0225 \). Seeing that it's positive, we confirm two real roots exist. These correspond to two points where the graph crosses the \(x\)-axis.

x-Intercepts

The \(x\)-intercepts of a quadratic function are the points where the graph intersects the \(x\)-axis. These points are important, as they can indicate the solutions to the related quadratic equation.
An \(x\)-intercept occurs where \(y=0\) in the function \( y = ax^2 + bx + c \).
In our problem, we used the discriminant to determine that the quadratic has two distinct \(x\)-intercepts because the discriminant was positive. To locate these intercepts, we found the roots using the quadratic formula.The coordinates of these \(x\)-intercepts, therefore, were calculated to be \((-3.40, 0)\) and \((-3.55, 0)\).Knowing the \(x\)-intercepts is useful for understanding the graph's behavior and can aid in sketching the graph accurately, as they tell us where the curve cuts through the \(x\)-axis.

Quadratic Formula

The quadratic formula is a reliable method to find the roots of any quadratic equation \( ax^2 + bx + c = 0 \). This formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula offers a systematic way to calculate the roots (or x-intercepts) based on the coefficients \(a\), \(b\), and \(c\) from the quadratic equation, without needing to graph the function.
- The "\( \pm \)" symbol shows there are typically two solutions: one using the '+' sign and another using the '−' sign.For the equation \( y = x^2 + 6.95x + 12.07 \), we calculated the discriminant first (\(d = 0.0225\)) and then applied these values into the quadratic formula:
\[ x_1, x_2 = \frac{-6.95 \pm \sqrt{0.0225}}{2 \times 1} \]By solving, we found the exact \(x\)-intercepts to be \( x_1 = -3.40 \) and \( x_2 = -3.55 \).
These roots represent the points at which the parabola crosses the \(x\)-axis. The quadratic formula is particularly handy because it also considers complex roots when the discriminant is negative. Thus, it's a versatile tool in solving quadratic equations.