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Chapter 8: Problem 44

Find the vertex and intercepts for each quadratic function. Sketch the graph, and state the domain and range. $$f(x)=x^{2}+2 x-3$$

Short Answer

Expert verified
Vertex: (-1, -4). Intercepts: x-intercepts (-3, 0) and (1, 0), y-intercept: (0, -3). Domain: (-鈭�, 鈭�). Range: [-4, 鈭�).

Step by step solution

01

- Find the Vertex

The vertex of a quadratic function in standard form, \(ax^2 + bx + c\), is given by the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = 2\), and \(c = -3\). Thus, the x-coordinate of the vertex is \(x = -\frac{2}{2(1)} = -1\). To find the y-coordinate, substitute \(x = -1\) back into the function: \(f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4\). Therefore, the vertex is \((-1, -4)\).
02

- Find the X-Intercepts

The x-intercepts occur where \(f(x) = 0\). To find these, solve the equation \(x^2 + 2x - 3 = 0\). Factoring the quadratic equation, we get \((x + 3)(x - 1) = 0\). Thus, the solutions are \(x = -3\) and \(x = 1\). Therefore, the x-intercepts are \((-3, 0)\) and \((1, 0)\).
03

- Find the Y-Intercept

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function: \(f(0) = 0^2 + 2(0) - 3 = -3\). Therefore, the y-intercept is \((0, -3)\).
04

- Sketch the Graph

Using the vertex \((-1, -4)\), the x-intercepts \((-3, 0)\) and \((1, 0)\), and the y-intercept \((0, -3)\), plot these points on a coordinate plane. Draw a smooth parabolic curve through these points, opening upwards since the coefficient of \(x^2\) is positive.
05

- State the Domain and Range

The domain of any quadratic function is all real numbers, represented as \(( -\infty, \infty )\). The range of this function, given that it opens upwards and has a vertex at \((-1, -4)\), starts at the y-coordinate of the vertex and goes to positive infinity. Therefore, the range is \([-4, \infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex Form
The vertex form of a quadratic function is a helpful way to quickly determine the vertex of the function. The vertex form is given by: \(f(x) = a(x-h)^2 + k\)
Here,
  • \(a\) determines the direction (upwards or downwards) and width of the parabola.
  • \(h\) and \(k\) are the coordinates of the vertex \((h, k)\).
To convert our function \(f(x) = x^2 + 2x - 3\) into vertex form, follow these steps:
1. Rewrite it in completed square form:\(f(x) = (x+1)^2-4\)2. Now we can easily identify the vertex to be at \((-1, -4)\).
This method is particularly useful when sketching or analyzing graphs, as it gives the vertex directly.
Intercepts
Understanding intercepts is crucial for graphing quadratic functions.
X-Intercepts: These are the points where the graph crosses the x-axis. They are found by setting \(f(x) = 0\) and solving for \(x\).
For our function \(f(x) = x^2 + 2x - 3\):
1. Set the equation to zero and factorize: \(x^2 + 2x - 3 = 0\)
2. Factor to get: \((x + 3)(x - 1) = 0\)3. Solving this gives the x-intercepts at (-3, 0) and (1, 0).
Y-Intercept: This is the point where the graph crosses the y-axis. It is found by setting \(x = 0\) and solving for \(y\).
For our function \(f(x) = x^2 + 2x - 3\):
1. Substitute 0 for \(x\):\(f(0) = 0^2 + 2(0) - 3 = -3\)Thus, the y-intercept is (0, -3).
Graphing Quadratics
Graphing a quadratic function involves plotting points and understanding the shape of the parabola. To graph our function \(f(x) = x^2 + 2x - 3\):
1. Start with the vertex, which is \((-1, -4)\).2. Plot the x-intercepts, at \((-3, 0)\) and (1, 0).3. Plot the y-intercept, at (0, -3).4. Draw the parabola through these points, ensuring it opens upwards since the coefficient of \(x^2\) is positive.
Additionally, identify the axis of symmetry. For our function, it is the vertical line passing through the vertex, \(x = -1\).Plot extra points to ensure accuracy, especially for a precise curve.
Domain and Range
Understanding the domain and range of a quadratic function is fundamental.
Domain: This refers to all possible values of \(x\). For any quadratic function, the domain is all real numbers. So, for \(f(x) = x^2 + 2x - 3\), the domain is ( -鈭�, 鈭� ).
Range: This refers to all possible values of \(y\). Since our quadratic function opens upwards and has a vertex at \((-1, -4)\), the smallest value of \(y\) is -4. Thus, the range of the function is [ -4, 鈭� ), as \(y\) can take any value greater than or equal to -4.By understanding these concepts, you can better interpret and graph quadratic functions.

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