1. **Problem statement:** We are given the function $f(x) = -2x^2 + 3x$ for $x \leq 1$ and need to solve for the point $B(1, 2)$ and the line $C_f$ with slope $\lambda = 1$ at $x_0 = 1$. 2. **Step 1: Evaluate $f(1)$** Calculate the value of the function at $x=1$: $$f(1) = -2(1)^2 + 3(1) = -2 + 3 = 1$$ So, the point $A$ on the curve is $A(1, 1)$. 3. **Step 2: Find the derivative $f'(x)$** The derivative of $f(x)$ is: $$f'(x) = \frac{d}{dx}(-2x^2 + 3x) = -4x + 3$$ 4. **Step 3: Calculate the slope at $x=1$** Evaluate $f'(1)$: $$f'(1) = -4(1) + 3 = -4 + 3 = -1$$ 5. **Step 4: Equation of the tangent line $C_f$ at $x=1$** The tangent line at $x=1$ has slope $f'(1) = -1$ and passes through $A(1,1)$. Using point-slope form: $$y - 1 = -1(x - 1)$$ Simplify: $$y - 1 = -x + 1$$ $$y = -x + 2$$ 6. **Step 5: Check the given slope $\lambda = 1$** The problem states $C_f$ has slope $\lambda = 1$, but the derivative at $x=1$ is $-1$, so the line with slope 1 is not tangent to $f$ at $x=1$. 7. **Step 6: Equation of line with slope 1 through $B(1,2)$** Using point-slope form: $$y - 2 = 1(x - 1)$$ Simplify: $$y - 2 = x - 1$$ $$y = x + 1$$ **Final answers:** - $f(1) = 1$, so $A(1,1)$ is on the curve. - Tangent line at $x=1$ is $y = -x + 2$ with slope $-1$. - Line with slope $1$ through $B(1,2)$ is $y = x + 1$. Since the problem asks to solve for $b4$ (likely the line with slope 1), the equation is: $$\boxed{y = x + 1}$$