1. The problem states we have a quadratic function $f(x) = 3x^2 + 4x - 5$ and a straight line passing through the point $(1,-1)$ with gradient $m$. The line is described by the equation $y = mx + c$. 2. Since the line passes through $(1,-1)$, substitute to find $c$: $$-1 = m\times 1 + c \implies c = -1 - m$$ So the line equation is: $$y = mx - 1 - m$$ 3. The line does not intersect the curve if the quadratic equation formed by equating $f(x)$ and the line has no real solution. Set $f(x) = y$: $$3x^2 + 4x - 5 = mx - 1 - m$$ Bring all terms to one side: $$3x^2 + 4x - 5 - mx + 1 + m = 0$$ Simplify: $$3x^2 + (4 - m)x + (m - 4) = 0$$ 4. For no intersection, the discriminant must be less than 0: $$\Delta = b^2 - 4ac < 0$$ Here, $a=3$, $b=4 - m$, and $c=m - 4$. Calculate discriminant: $$\Delta = (4 - m)^2 - 4 \times 3 \times (m - 4) < 0$$ 5. Expand and simplify: $$(4 - m)^2 = 16 - 8m + m^2$$ So: $$16 - 8m + m^2 - 12m + 48 < 0$$ Simplify: $$m^2 - 20m + 64 < 0$$ 6. Solve the quadratic inequality: Find roots: $$m = \frac{20 \pm \sqrt{400 - 256}}{2} = \frac{20 \pm \sqrt{144}}{2} = \frac{20 \pm 12}{2}$$ So roots are: $$m = 4 \quad \text{and} \quad m = 16$$ 7. Since $a=1 > 0$, the quadratic opens upward, so $m^2 - 20m + 64 < 0$ between roots: $$4 < m < 16$$ Final answer: The gradient $m$ of the straight line must lie in the open interval $$\boxed{(4, 16)}$$ so that the line does not intersect the curve.