1. **Problem statement:** Graph the function $y=3x^2$ and find the equation of the tangent line to the curve at the point where $x=1$. 2. **Formula and rules:** The slope of the tangent line to a curve $y=f(x)$ at $x=a$ is given by the derivative $f'(a)$. The tangent line equation at $x=a$ is: $$y = f(a) + f'(a)(x - a)$$ 3. **Find the derivative:** For $y=3x^2$, the derivative is: $$y' = \frac{d}{dx}(3x^2) = 6x$$ 4. **Evaluate the derivative at $x=1$:** $$y'(1) = 6 \times 1 = 6$$ This is the slope of the tangent line at $x=1$. 5. **Find the point on the curve at $x=1$:** $$y(1) = 3 \times 1^2 = 3$$ So the point is $(1,3)$. 6. **Write the tangent line equation:** Using point-slope form: $$y = 3 + 6(x - 1) = 6x - 3$$ 7. **Summary:** The parabola $y=3x^2$ opens upwards with vertex at $(0,0)$. The tangent line at $x=1$ passes through $(1,3)$ with slope $6$, equation $y=6x-3$. This matches graph B, where the tangent line has a positive slope larger than the curve at that point.