Problem 14 - Beginning and Intermediate Algebra

Question

Find the vertex of each parabola. For each equation, decide whether the graph opens up, down, to the left, or to the right, and whether it is wider, narrower, or the same shape as the graph of \(y=x^{2}\). If it is a parabola with a vertical axis of symmetry, find the discriminant and use it to determine the number of \(x\) -intercepts. $$ f(x)=3 x^{2}-6 x+4 $$

Step-by-Step Solution

Answer

Vertex: (1, 1). The parabola opens upwards, is narrower than y=x², and has no x-intercepts.

Step 1: Identify the coefficients

Rewrite the equation in the standard form of a quadratic equation: \[ f(x) = ax^2 + bx + c \]For the given equation, \[ f(x) = 3x^2 - 6x + 4 \]here, \[ a = 3, \, b = -6, \, c = 4 \]

Step 2: Find the vertex

The vertex of the parabola can be found using the formula for the x-coordinate of the vertex: \[ x = \frac{-b}{2a} \]Substitute the values of \( a \) and \( b \): \[ x = \frac{-(-6)}{2(3)} = \frac{6}{6} = 1 \]Now, substitute \( x = 1 \) back into the original equation to find the y-coordinate: \[ y = 3(1)^2 - 6(1) + 4 = 3 - 6 + 4 = 1 \]Therefore, the vertex of the parabola is \[ (1, 1) \]

Step 3: Determine the direction of the graph

The coefficient \( a \) determines the direction the parabola opens. If \( a > 0 \), the parabola opens upwards. If \( a < 0 \), the parabola opens downwards. Since \( a = 3 > 0 \), the parabola opens upwards.

Step 4: Determine the width of the graph

Compare the coefficient \( a \) with 1. If \( |a| > 1 \), the parabola is narrower than \( y = x^2 \). If \( |a| < 1 \), the parabola is wider. Since \( 3 > 1 \), the graph is narrower than \( y = x^2 \).

Step 5: Find the discriminant

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c \) can be determined using the formula: \[ \Delta = b^2 - 4ac \]Substitute the values of \( a \), \( b \), and \( c \): \[ \Delta = (-6)^2 - 4(3)(4) = 36 - 48 = -12 \]

Step 6: Determine the number of x-intercepts

The sign of the discriminant \( \Delta \) determines the number of x-intercepts of the quadratic equation: - If \( \Delta > 0 \), there are 2 distinct real x-intercepts. - If \( \Delta = 0 \), there is 1 real x-intercept (a double root). - If \( \Delta < 0 \), there are no real x-intercepts. Since \( \Delta = -12 < 0 \), there are no real x-intercepts.

Key Concepts

vertex of a parabola

The vertex is a key feature of a parabola. It's the highest or lowest point on the graph, depending on its direction. For the quadratic equation\(f(x) = 3x^2 - 6x + 4\), we determined the vertex using the vertex formula: \(x = \frac{-b}{2a}\). By substituting the values from our equation, we found that the x-coordinate of the vertex is 1.
Next, we substituted back into the original equation to find the y-coordinate, which is 1. Therefore, the vertex of the parabola is at \((1, 1)\). This point is important because it helps us understand the shape and position of the graph on the coordinate plane.

quadratic equation

A quadratic equation is a second-degree polynomial that can be written in the standard form: \(ax^2 + bx + c\). In our example, \(f(x) = 3x^2 - 6x + 4\), the coefficients are \(a = 3\), \(b = -6\), and \(c = 4\).
Quadratic equations are important because they model many real-world scenarios like projectile motion and areas. By analyzing the coefficients, we can determine several properties of the graph of the quadratic function.

discriminant

The discriminant is a specific value calculated from the coefficients of a quadratic equation \(ax^2 + bx + c\). It is given by the formula \(\Delta = b^2 - 4ac\). In our case, substituting \(a = 3\), \(b = -6\), and \(c = 4\) into the formula, we get \(\Delta = 36 - 48 = -12\).
The discriminant determines the number of real x-intercepts:

• If \(\Delta > 0\), the equation has two distinct real solutions.
• If \(\Delta = 0\), it has one real solution.
• If \(\Delta < 0\), there are no real solutions, just like in our example where \(\Delta = -12\).

The fact that \(\Delta < 0\) means our parabola does not cross the x-axis.

graph direction

The direction of the graph of a quadratic equation depends on the coefficient \(a\). If \(a > 0\), the parabola opens upwards, and if \(a < 0\), it opens downwards. In our example, \(a = 3\), which is greater than 0, so the parabola opens upwards.
Understanding the direction is crucial for predicting the parabola's behavior. Since our parabola opens upwards and the vertex is at \((1, 1)\), we know it has a minimum point at this vertex.

graph width

The width of the parabola is determined by the absolute value of \(a\). If \(|a| > 1\), the graph is narrower than the basic parabola \(y = x^2\). Conversely, if \(|a| < 1\), the graph is wider.
In our case, \(a = 3\), and since \(3 > 1\), the graph of our quadratic equation is narrower than \(y = x^2\). This means the parabola will be steeper, and its sides will close in more sharply.

x-intercepts

X-intercepts are the points where the graph crosses the x-axis. For quadratic equations, the number of x-intercepts can be determined using the discriminant \(\Delta\). As mentioned earlier, if \(\Delta > 0\), there are two x-intercepts; if \(\Delta = 0\), there is one; and if \(\Delta < 0\), there are none.
For our equation \(f(x) = 3x^2 - 6x + 4\), we found that \(\Delta = -12\), which is less than 0. Therefore, this quadratic equation has no real x-intercepts. This also means the graph stays above the x-axis or below if the parabola opens downwards.