A step-by-step tutorial on graphing and sketching square root functions. The graph, domain, range, algebraic simplifications, and other properties are discussed.
The domain of function \( f \) defined by \( f(x) = \sqrt{x} \) is the set of all real positive numbers and zero, because square roots of negative numbers are not real numbers (e.g., \( \sqrt{-4} \) is not real).
Problem: Make a table of values for function \( f(x) = \sqrt{x} \), graph it, and find its range.
Solution:
Because the domain of \( f \) is the set of all positive real numbers and zero, we construct a table of values with inputs whose square roots are whole numbers:
| \( x \) | \(0\) | \(1\) | \(4\) | \(9\) | \(16\) |
| \( \sqrt{x} \) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
These values make it easy to plot the reference points.
Problem: Find the domain, make a table of values for \( f(x) = \sqrt{x - 3} \), graph it, and find its range.
Solution:
Set the expression under the square root greater than or equal to zero: \[ x - 3 \ge 0 \implies x \ge 3 \] Selecting values in the domain:
| \( x \) | \(3\) | \(4\) | \(7\) | \(12\) |
| \( \sqrt{x - 3} \) | \(0\) | \(1\) | \(2\) | \(3\) |
Problem: Find the domain, construct a table of values for \( f(x) = -\sqrt{-2x + 4} + 1 \), graph it, and find its range.
Solution:
Set the radicand greater than or equal to zero: \[ -2x + 4 \ge 0 \implies -2x \ge -4 \implies x \le 2 \] *(Note: Original text stated \( x \ge 2 \), but solving \(-2x + 4 \ge 0\) yields \( x \le 2 \). Let's use valid domain points accordingly:)*
| \( x \) | \(2\) | \(\dfrac{3}{2}\) | \(0\) | \(-\dfrac{5}{2}\) | \(-6\) |
| \( -\sqrt{-2x + 4} + 1 \) | \(1\) | \(0\) | \(-1\) | \(-2\) | \(-3\) |
Problem: Find the domain of \( f(x) = \sqrt{-x^2 + 4} \), graph it, and find its range.
Solution:
Solve the polynomial inequality: \[ -x^2 + 4 \ge 0 \implies x^2 \le 4 \implies -2 \le x \le 2 \] The domain is interval \( [-2, 2] \). Rewriting as an equation: \[ y = \sqrt{-x^2 + 4} \implies y^2 = -x^2 + 4 \implies x^2 + y^2 = 2^2 \] This is the equation of a circle. Because the principal square root is non-negative, the graph is the upper half of the circle.
Problem: Find the domain, make a table of values for \( f(x) = \sqrt{x^2 - 9} \), graph it, and find its range.
Solution:
Solve \( x^2 - 9 \ge 0 \), which gives the domain: \[ (-\infty, -3] \cup [3, +\infty) \] Since \( f(x) = f(-x) \), the graph is symmetric with respect to the \( y \)-axis.
| \( x \) | \(3\) | \(5\) | \(8\) |
| \( \sqrt{x^2 - 9} \) | \(0\) | \(4\) | \(7.4\) |
Problem: Simplify \( f(x) = \sqrt{x^2 - 6x + 9} \), graph it, and find its range.
Solution:
Rewrite the expression under the square root as a perfect square trinomial: \[ x^2 - 6x + 9 = (x - 3)^2 \] Hence: \[ f(x) = \sqrt{(x - 3)^2} = |x - 3| \] The function simplifies to an absolute value function, which can be graphed as a V-shaped piecewise curve.
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