1. **State the problem:** We are given a straight line when plotting $e^y$ against $x^2$, with gradient $-3$ passing through $(4.30,5.85)$. We need to: (a) Find $y$ in terms of $x$. (b) Find the values of $x$ for which $y$ exists. 2. **Given:** Define $X = x^2$ and $Y = e^y$. The line equation is $$Y - 5.85 = -3(X - 4.30)$$ Simplify it: $$Y = -3X + 12.9 + 5.85 = -3x^2 + 18.75$$ so $$e^y = -3x^2 + 18.75$$ 3. **Find $y$ in terms of $x$: ** Take natural log on both sides assuming $e^y > 0$: $$y = \\ln\left(e^y\right) = \\ln\left(-3x^2 + 18.75\right)$$ Hence, $$y = \ln\left(18.75 - 3x^2\right)$$ 4. **Find values of $x$ for which $y$ exists:** Since $y$ is defined only if the argument of the $\\ln$ is positive, $$18.75 - 3x^2 > 0$$ Divide both sides by 3: $$6.25 - x^2 > 0$$ Rewrite as: $$x^2 < 6.25$$ Take square roots: $$-\sqrt{6.25} < x < \sqrt{6.25}$$ $$-2.5 < x < 2.5$$ **Final answers:** (a) $$y = \ln\left(18.75 - 3x^2\right)$$ (b) $$x \in (-2.5, 2.5)$$