Problem 27 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=-\left(x^{2}+2 x-3\right)$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (1,0). The intercepts are at (1,0), (-3,0) for x-intercepts and (0,3) for the y-intercept.

Step 1: Identify the Vertex

The vertex of the parabola is the point \(v(h, k)\) where \(h = -\frac{b}{2a}\) and \(k = f(h)\). For the function \(f(x) =- (x^{2} +2x -3)\), \(a = -1\), \(b =2\), and \(c = -3\). So, \(h = - \frac{2}{2*(-1)} = 1\), and the y-coordinate \(k = f(1) = -(1^2 + 2*1 - 3) = 0\). So the vertex of the parabola is at \(v(1, 0)\).

Step 2: Identify the Intercepts

The x-intercepts can be found by setting \(f(x) = 0\) and solving for \(x\). This gives us the roots of the quadratic. \(0=-(x^{2} +2x -3)\) becomes \(0=x^{2} +2x -3\). Factoring out, we get \(0=(x-1)(x+3)\), so the roots are \(x=1\) and \(x=-3\). These are also our x-intercepts (1,0) and (-3,0). The y-intercept is found by setting \(x=0\) in the function, i.e., \(f(0)= - (0 + 0 - 3) = 3\). So, the y-intercept is (0,3).

Step 3: Sketch the Graph

Plotting the vertex (1,0), the x-intercepts (1,0), (-3,0) and the y-intercept (0,3), and with the knowledge that the parabola opens downwards (since \(a = -1\) is negative), we can now sketch the parabola.

Key Concepts

Vertex of a Parabola

In a quadratic function like \( f(x) = ax^2 + bx + c \),the vertex represents the highest or lowest point of the parabola, depending on whether it opens upwards or downwards.
It can be found using the formula:

• \( h = -\frac{b}{2a}\) for the x-coordinate,
• and \( k = f(h) \) for the y-coordinate.

For the function \( f(x) = -(x^2 + 2x - 3) \),the values are \( a = -1 \), \( b = 2 \), and \( c = -3 \).By substituting these into the formula, we find that \( h = 1 \) and \( k = 0 \).
This makes the vertex \((1, 0)\). The vertex is important because it provides a starting point for graphing and helps us understand the parabola's direction.
In this case, the parabola opens downwards
due to the negative value of \( a \).

Intercepts of a Graph

Intercepts are points where the graph crosses the axes. They are crucial because they help us understand where the quadratic equation is equal to zero (x-intercepts) or what the function's value is when \( x \) is zero (y-intercept).

• X-intercepts: Set \( f(x) = 0 \) to find \( x \). Solve \( 0 = -(x^2 + 2x -3) \) to get the equation \( 0 = (x-1)(x+3) \). The solutions, \( x = 1 \) and \( x = -3 \), are the x-intercepts \((1, 0)\) and \((-3, 0)\).
• Y-intercept: Set \( x = 0 \) and compute \( f(0) \). For \( f(x) = -(x^2 + 2x -3) \), this becomes \( 3 \), so the y-intercept is \((0, 3)\).

These intercepts are vital markers on the graph, providing clues about its shape and orientation.
The x-intercepts reflect where the graph touches the x-axis, and the y-intercept shows where the graph intersects the y-axis.

Graphing Quadratic Equations

Graphing a quadratic equation like \( f(x) = -(x^2 + 2x -3) \) involves plotting key points such as the vertex and intercepts and recognizing the general shape of the curve.
This quadratic opens downward, evident from the negative leading coefficient \( a = -1 \).

• Start with the vertex: Plot the vertex \((1, 0)\). This is a guide for the parabola's center.
• Plot the intercepts: Include the x-intercepts \((1, 0)\) and \((-3, 0)\), and the y-intercept \((0, 3)\).
• Draw the curve: Use a smooth, symmetrical U-shape that connects these points.
  The parabola will go through the vertex and intercepts, forming a downward curve.

With these steps, you can visualize the parabolic curve, understanding how it is influenced by the equation's parameters. Recognizing patterns in the coefficients help predict how the graph will look. Practice is key to mastering this process, enabling you to read and sketch quadratics confidently.