1. **State the problem:** Find the equation of the line of symmetry of the curve for the quadratic function $$y = 2x^2 - 4x + 3$$. 2. **Recall the formula for the axis of symmetry:** For a quadratic function $$y = ax^2 + bx + c$$, the line of symmetry (axis of symmetry) is given by: $$x = -\frac{b}{2a}$$ 3. **Identify coefficients:** Here, $$a = 2$$ and $$b = -4$$. 4. **Calculate the line of symmetry:** $$x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1$$ 5. **Interpretation:** The line of symmetry is the vertical line $$x = 1$$. 1. **State the problem:** Find the gradient of the curve $$y = 2x^2 - 4x + 3$$ at $$x = 3$$. 2. **Recall the formula for gradient:** The gradient of a curve at a point is the derivative $$\frac{dy}{dx}$$ evaluated at that point. 3. **Differentiate the function:** $$\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 4x + 3) = 4x - 4$$ 4. **Evaluate at $$x=3$$:** $$\frac{dy}{dx}\bigg|_{x=3} = 4(3) - 4 = 12 - 4 = 8$$ 5. **Interpretation:** The gradient of the curve at $$x=3$$ is $$8$$. 1. **State the problem:** Find the minimum value of $$y = 2x^2 - 4x + 3$$. 2. **Recall:** For a quadratic $$ax^2 + bx + c$$ with $$a > 0$$, the minimum value occurs at $$x = -\frac{b}{2a}$$. 3. **Calculate $$x$$ for minimum:** As before, $$x = 1$$. 4. **Find minimum value by substituting $$x=1$$ into $$y$$:** $$y = 2(1)^2 - 4(1) + 3 = 2 - 4 + 3 = 1$$ 5. **Interpretation:** The minimum value of $$y$$ is $$1$$ at $$x=1$$.