1. **Problem 1:** Find the values of $h$, $m$, and $c$ for the function $f(q) = hq^2 + mq + c$ given the gradient function $4q + 8$ and a stationary value of $-3$. 2. **Problem 2:** Find the equation of the tangent line to the curve $y = -\frac{10}{x}$ at the point where $x = -5$. 3. **Problem 3:** Given $m(q) = 2qe^{-q^2}$: a. State the differentiation rules used and the parts of $m(q)$ they apply to. b. Find the derivative $m'(0)$. --- ### Step 1: Problem 1 1. The gradient function of $f(q)$ is the derivative $f'(q) = 2hq + m$. 2. We are given $f'(q) = 4q + 8$, so equate: $$2hq + m = 4q + 8$$ 3. Equate coefficients: $$2h = 4 \Rightarrow h = 2$$ $$m = 8$$ 4. The stationary point occurs where $f'(q) = 0$: $$4q + 8 = 0 \Rightarrow q = -2$$ 5. The stationary value is $f(-2) = -3$: $$f(-2) = h(-2)^2 + m(-2) + c = 2(4) + 8(-2) + c = 8 - 16 + c = -8 + c$$ 6. Set equal to $-3$: $$-8 + c = -3 \Rightarrow c = 5$$ **Answer for Problem 1:** $h=2$, $m=8$, $c=5$ --- ### Step 2: Problem 2 1. The curve is $y = -\frac{10}{x} = -10x^{-1}$. 2. Differentiate using the power rule: $$y' = -10 \times (-1) x^{-2} = 10x^{-2} = \frac{10}{x^2}$$ 3. Find the slope at $x = -5$: $$y'(-5) = \frac{10}{(-5)^2} = \frac{10}{25} = \frac{2}{5}$$ 4. Find the point on the curve at $x = -5$: $$y(-5) = -\frac{10}{-5} = 2$$ 5. Equation of tangent line at $(x_0, y_0)$ is: $$y - y_0 = m(x - x_0)$$ $$y - 2 = \frac{2}{5}(x + 5)$$ 6. Simplify: $$y = \frac{2}{5}x + 2 + 2 = \frac{2}{5}x + 4$$ **Answer for Problem 2:** $y = \frac{2}{5}x + 4$ --- ### Step 3: Problem 3 **a. Differentiation rules and parts:** - Product Rule: for $2q \cdot e^{-q^2}$ - Chain Rule: for $e^{-q^2}$ because the exponent is a function of $q$ **b. Find $m'(q)$:** 1. Using product rule: $$m'(q) = (2q)' e^{-q^2} + 2q (e^{-q^2})'$$ 2. Derivative of $2q$ is 2. 3. Derivative of $e^{-q^2}$ using chain rule: $$\frac{d}{dq} e^{-q^2} = e^{-q^2} \times (-2q) = -2q e^{-q^2}$$ 4. Substitute back: $$m'(q) = 2 e^{-q^2} + 2q (-2q e^{-q^2}) = 2 e^{-q^2} - 4q^2 e^{-q^2}$$ 5. Factor: $$m'(q) = e^{-q^2} (2 - 4q^2)$$ 6. Evaluate at $q=0$: $$m'(0) = e^{0} (2 - 0) = 1 \times 2 = 2$$ **Answer for Problem 3:** - a. Product and Chain rules - b. $m'(0) = 2$