1. **State the problem:** Find the slope of the tangent line to the graph of $f(x) = 3x^2 - 12x + 14$ at the point $(1,5)$ and then find the equation of the tangent line in the form $y = mx + b$. 2. **Formula for slope of tangent line:** The slope $m$ at $x=1$ is given by the derivative of $f(x)$ at $x=1$, which can be found using the limit definition: $$m = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$$ 3. **Calculate $f(1)$:** $$f(1) = 3(1)^2 - 12(1) + 14 = 3 - 12 + 14 = 5$$ 4. **Calculate $f(1+h)$:** $$f(1+h) = 3(1+h)^2 - 12(1+h) + 14 = 3(1 + 2h + h^2) - 12 - 12h + 14 = 3 + 6h + 3h^2 - 12 - 12h + 14$$ Simplify: $$= (3 - 12 + 14) + (6h - 12h) + 3h^2 = 5 - 6h + 3h^2$$ 5. **Form the difference quotient:** $$\frac{f(1+h) - f(1)}{h} = \frac{(5 - 6h + 3h^2) - 5}{h} = \frac{-6h + 3h^2}{h}$$ 6. **Simplify the fraction by canceling $h$:** $$\frac{\cancel{h}(-6 + 3h)}{\cancel{h}} = -6 + 3h$$ 7. **Take the limit as $h \to 0$:** $$m = \lim_{h \to 0} (-6 + 3h) = -6 + 3(0) = -6$$ 8. **Find the equation of the tangent line:** We know the slope $m = -6$ and the point $(1,5)$ lies on the line. Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 5 = -6(x - 1)$$ 9. **Simplify to slope-intercept form $y = mx + b$:** $$y - 5 = -6x + 6$$ $$y = -6x + 6 + 5$$ $$y = -6x + 11$$ 10. **Identify $m$ and $b$:** $$m = -6$$ $$b = 11$$ **Final answer:** The slope of the tangent line at $(1,5)$ is $-6$. The equation of the tangent line is $$y = -6x + 11$$.