Problem 3 - Precalculus

Question

In Exercises \(1-12\), sketch the graph of the given function. State the domain of the function, identify any intercepts and test for symmetry. $$ f(x)=x^{2}+1 $$

Step-by-Step Solution

Answer

The graph of \(f(x) = x^2 + 1\) is a parabola opening upward with vertex (0, 1), domain \((-\infty, \infty)\), y-intercept (0,1), no x-intercepts, and is symmetric about the y-axis.

Step 1: Identify the Domain of the Function

The domain of a function is the set of all possible input values (usually x-values) that the function can accept. Since the function is a polynomial, specifically a quadratic function, there are no restrictions on the x-values it can take. Therefore, the domain is all real numbers, written as \((-\infty, \infty)\).

Step 2: Sketch the Graph of the Function

To sketch the graph of \(f(x) = x^2 + 1\), recognize that it is a transformation of the parent function \(y = x^2\). The graph of \(y = x^2\) is a parabola opening upwards with vertex at the origin. For \(f(x) = x^2 + 1\), the entire graph is shifted up by 1 unit. So, the vertex of the parabola is now at (0, 1). The general shape remains the same, opening upwards.

Step 3: Identify Intercepts

The y-intercept is found by setting \(x = 0\): \(f(0) = 0^2 + 1 = 1\). So, the y-intercept is at (0, 1). For x-intercepts, set \(f(x) = 0\), which yields the equation \(x^2 + 1 = 0\). Solving for \(x\), we find this equation has no real solutions (as \(x^2\) cannot be negative). Thus, there are no x-intercepts.

Step 4: Test for Symmetry

A function is symmetric about the y-axis if \(f(-x) = f(x)\). Compute \(f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)\). This confirms symmetry about the y-axis. It does not exhibit symmetry about the x-axis or origin since \(f(x) eq -f(x)\) and \(f(-x) eq -f(x)\). Therefore, the only symmetry is y-axis symmetry.

Key Concepts

Domain of Quadratic Functions

The domain of a function is the complete set of possible values of the variable, typically represented by \(x\), that can be input into the function. For quadratic functions, such as the example \(f(x) = x^2 + 1\), there are no restrictions on the values \(x\) can take. This is because quadratic functions are polynomials. Polynomials are defined for all real numbers. Therefore, the domain is all real numbers, which you can write as \((-infty, infty)\). This universal domain holds true for all quadratic functions unless specifically limited by context or other conditions like square roots or divisions by zero.

Intercepts of Quadratic Functions

Intercepts are points where the graph of a function crosses the axes. For a quadratic function like \(f(x) = x^2 + 1\), identifying intercepts helps understand its positioning on the graph.

• y-Intercept: To find the y-intercept, you set \(x = 0\). Plugging into the function, \(f(0) = 0^2 + 1\), gives us the y-intercept at the point \((0, 1)\). This point indicates where the parabola intersects the y-axis.
• x-Intercept(s): To find x-intercepts, you set \(f(x) = 0\), which leads to the equation \(x^2 + 1 = 0\). Solving it involves assuming \(x^2 = -1\), which yields no real solutions since squares of real numbers cannot be negative. Consequently, this quadratic function lacks genuine x-intercepts.

Understanding intercepts is crucial for sketching the overall graph of the function.

Symmetry of Functions

Quadratic functions often display symmetry, which simplifies their analysis and graphing. A function is symmetric about the y-axis if its equation remains unchanged when replacing \(x\) with \(-x\). For \(f(x) = x^2 + 1\), compute \(f(-x) = (-x)^2 + 1 = x^2 + 1\). Since the form is identical to the original, the function is symmetric about the y-axis.
Symmetry about the y-axis often means that for every point \((x, y)\), there is a corresponding point \((-x, y)\). This symmetry produces a predictable, mirror-like quality which allows us to understand and replicate the graph on either side of the y-axis. Be aware, though, that this function doesn't exhibit symmetry about the x-axis or origin, meanings where \(f(-x) = -f(x)\) or \(f(x) = -f(x)\). Only y-axis symmetry is present here.

Parabola Transformations

Transformations shift or adjust a parent graph to a new position. With the parent function \(y = x^2\), which is a standard parabola centered at the origin, transformations allow us to model more complex situations.

• Vertical Shifts: In \(f(x) = x^2 + 1\), the \(+1\) shifts the entire parabola up one unit. This means the vertex of the parabola moves from \((0, 0)\) to \((0, 1)\), resulting in its graph starting at \(y = 1\) instead of the origin.
• Horizontal Shifts: These occur if the form inside the function changes, like \(f(x) = (x-b)^2\) but are not present in our case.

Such transformations are vital in applying quadratic functions to real-world problems, allowing for adaptations to various initial conditions or models.