1. **State the problem:** We have a plot of $e^y$ versus $x^2$ that forms a straight line with gradient $-3$. The line passes through the point $(4.30, 5.85)$. We need to find (a) $y$ in terms of $x$ and (b) the values of $x$ for which $y$ exists. 2. **Express the line equation:** Since $e^y$ is plotted against $x^2$, let $X = x^2$ and $Y = e^y$. The line equation has gradient $m = -3$ and passes through $(X_1, Y_1) = (4.30, 5.85)$. The linear equation is: $$ Y - 5.85 = -3 (X - 4.30) $$ Simplify: $$ Y = -3X + 3 \times 4.30 + 5.85 = -3X + 12.9 + 5.85 = -3X + 18.75 $$ So, $$ e^y = -3x^2 + 18.75 $$ 3. **Find $y$ in terms of $x$:** Take the natural logarithm on both sides: $$ y = \ln\left(-3x^2 + 18.75\right) $$ 4. **Determine values of $x$ where $y$ exists:** The argument of the logarithm must be positive: $$ -3x^2 + 18.75 > 0 $$ Rearranged: $$ -3x^2 > -18.75 $$ Divide both sides by $-3$ (reverse inequality sign): $$ x^2 < \frac{18.75}{3} = 6.25 $$ So, $$ |x| < 2.5 $$ Therefore, $y$ exists only for $x$ in the interval $(-2.5, 2.5)$. **Final answers:** (a) $$ y = \ln\left(-3x^{2} + 18.75 \right) $$ (b) $$ -2.5 < x < 2.5 $$