1Cademy - Finding the Domain and Points on the Graph of $$f(x) = \frac{8-x}{x^2-7x+12}$$ when $$f(x) = 3$$

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Strategy to Solve Equations with Rational Expressions

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Finding the Domain and Points on the Graph of $f(x) = \frac{8-x}{x^2-7x+12}$ when $f(x) = 3$

To analyze the rational function $f(x) = \frac{8-x}{x^2-7x+12}$ , we determine its domain, solve the equation $f(x) = 3$ , and find the corresponding point on the graph.

Step 1: Find the domain. Set the denominator to zero to find the restricted values: $x^2 - 7x + 12 = 0$ $(x - 3)(x - 4) = 0$ $x - 3 = 0 \quad \text{or} \quad x - 4 = 0$ $x = 3 \quad \text{or} \quad x = 4$ The domain is all real numbers except $x \neq 3$ and $x \neq 4$ .

Step 2: Solve $f(x) = 3$ . Set the rational expression equal to $3$ : $\frac{8-x}{x^2-7x+12} = 3$ Factor the denominator to identify the LCD, which is $(x-3)(x-4)$ : $\frac{8-x}{(x-3)(x-4)} = 3$ Multiply both sides by the LCD to clear fractions: $(x-3)(x-4) \cdot \frac{8-x}{(x-3)(x-4)} = 3(x-3)(x-4)$ Simplify and distribute: $8 - x = 3(x^2 - 7x + 12)$ $8 - x = 3x^2 - 21x + 36$ Set the equation to zero by adding $x$ and subtracting $8$ from both sides: $0 = 3x^2 - 20x + 28$ Factor the quadratic equation: $0 = (3x - 14)(x - 2)$ Set each factor to zero to solve for $x$ : $3x - 14 = 0 \quad \text{or} \quad x - 2 = 0$ $x = \frac{14}{3} \quad \text{or} \quad x = 2$ Check against the domain restrictions ( $x \neq 3$ , $x \neq 4$ ). Neither solution is restricted, so both $x = \frac{14}{3}$ and $x = 2$ are valid solutions.

Step 3: Find the points on the graph. For both $x = \frac{14}{3}$ and $x = 2$ , the function value is $f(x) = 3$ . Therefore, the points $\left(\frac{14}{3}, 3\right)$ and (2, 3) lie on the graph of the function.

Updated 2026-06-18

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OpenStax Intermediate Algebra

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