1. **Problem statement:** (a) Given the function $y = 1 - 3e^{-x^2}$, we need to: (i) Calculate $y$ at $x=1$. (ii) Find the $x$-intercepts. (iii) Find the equation of the tangent line at $x=1$. (iv) Sketch the derivative $\frac{dy}{dx}$. (b) For a quadratic $f(x) = x^2 + bx + c$ with turning point at $\left(\frac{1}{2}, -\frac{49}{4}\right)$, find $b$ and $c$ and sketch the graph. --- 2. **Formulas and rules:** - To find $y$ at a point, substitute $x$ into the function. - $x$-intercepts occur where $y=0$. - The derivative $\frac{dy}{dx}$ gives slope of tangent. - Equation of tangent at $x=a$ is $y = y(a) + y'(a)(x - a)$. - For quadratic $f(x) = x^2 + bx + c$, vertex $x$-coordinate is $-\frac{b}{2}$. --- 3. **Step-by-step solutions:** **(a)(i) Calculate $y$ at $x=1$:** $$y = 1 - 3e^{-1^2} = 1 - 3e^{-1} = 1 - 3\times \frac{1}{e} = 1 - \frac{3}{e}$$ Numerically, $e \approx 2.718$, so $$y \approx 1 - \frac{3}{2.718} = 1 - 1.1036 = -0.1036$$ **(a)(ii) Find $x$-intercepts where $y=0$:** Set $$0 = 1 - 3e^{-x^2} \implies 3e^{-x^2} = 1 \implies e^{-x^2} = \frac{1}{3}$$ Take natural log: $$-x^2 = \ln\left(\frac{1}{3}\right) = -\ln(3) \implies x^2 = \ln(3)$$ So, $$x = \pm \sqrt{\ln(3)}$$ Numerically, $\ln(3) \approx 1.0986$, so $$x \approx \pm 1.048$$ **(a)(iii) Equation of tangent at $x=1$:** First find derivative: $$y = 1 - 3e^{-x^2}$$ $$\frac{dy}{dx} = -3 \times \frac{d}{dx} e^{-x^2} = -3 \times e^{-x^2} \times (-2x) = 6xe^{-x^2}$$ At $x=1$: $$y'(1) = 6 \times 1 \times e^{-1} = 6 \times \frac{1}{e} = \frac{6}{e} \approx 2.207$$ Recall from (i): $$y(1) = 1 - \frac{3}{e} \approx -0.1036$$ Equation of tangent: $$y = y(1) + y'(1)(x - 1) = -0.1036 + 2.207(x - 1)$$ Or in exact form: $$y = 1 - \frac{3}{e} + \frac{6}{e}(x - 1)$$ **(a)(iv) Sketch derivative $\frac{dy}{dx}$:** - Derivative is $6xe^{-x^2}$. - Note $e^{-x^2} > 0$ always. - Derivative is zero at $x=0$. - For $x>0$, derivative positive; for $x<0$, derivative negative. - Derivative has a bell shape multiplied by $x$, so it crosses zero at $x=0$ and tends to zero as $x \to \pm \infty$. **(b) Quadratic with turning point at $\left(\frac{1}{2}, -\frac{49}{4}\right)$:** General form: $$f(x) = x^2 + bx + c$$ Vertex $x$-coordinate: $$x_v = -\frac{b}{2} = \frac{1}{2} \implies b = -1$$ Vertex $y$-coordinate: $$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + b \times \frac{1}{2} + c = -\frac{49}{4}$$ Substitute $b = -1$: $$\frac{1}{4} - \frac{1}{2} + c = -\frac{49}{4}$$ Simplify left side: $$-\frac{1}{4} + c = -\frac{49}{4} \implies c = -\frac{49}{4} + \frac{1}{4} = -\frac{48}{4} = -12$$ So, $$b = -1, \quad c = -12$$ **Sketch features:** - Vertex at $\left(\frac{1}{2}, -\frac{49}{4}\right)$ - Parabola opens upward (coefficient of $x^2$ is positive) - Axis of symmetry $x = \frac{1}{2}$ - $y$-intercept at $f(0) = c = -12$ --- **Final answers:** (i) $y(1) = 1 - \frac{3}{e} \approx -0.1036$ (ii) $x$-intercepts at $x = \pm \sqrt{\ln(3)} \approx \pm 1.048$ (iii) Tangent line at $x=1$: $$y = 1 - \frac{3}{e} + \frac{6}{e}(x - 1)$$ (iv) Derivative function: $$\frac{dy}{dx} = 6xe^{-x^2}$$ (b) Quadratic coefficients: $$b = -1, \quad c = -12$$