1. The problem asks which quadratic function has solutions at $1$ and $-\frac{3}{4}$. 2. Recall that if a quadratic function is factored as $(x - r_1)(x - r_2) = 0$, then the solutions (roots) are $r_1$ and $r_2$. 3. We want to find the function whose roots are $1$ and $-\frac{3}{4}$. 4. Check each option by setting each factor equal to zero and solving for $x$: - Option A: $(x + 1)(4x - 3) = 0$ - $x + 1 = 0 \Rightarrow x = -1$ - $4x - 3 = 0 \Rightarrow x = \frac{3}{4}$ - Roots are $-1$ and $\frac{3}{4}$, which do not match. - Option B: $(x + 1)(4x + 3) = 0$ - $x + 1 = 0 \Rightarrow x = -1$ - $4x + 3 = 0 \Rightarrow x = -\frac{3}{4}$ - Roots are $-1$ and $-\frac{3}{4}$, which do not match. - Option C: $(x - 1)(4x - 3) = 0$ - $x - 1 = 0 \Rightarrow x = 1$ - $4x - 3 = 0 \Rightarrow x = \frac{3}{4}$ - Roots are $1$ and $\frac{3}{4}$, which do not match. - Option D: $(x - 1)(4x + 3) = 0$ - $x - 1 = 0 \Rightarrow x = 1$ - $4x + 3 = 0 \Rightarrow x = -\frac{3}{4}$ - Roots are $1$ and $-\frac{3}{4}$, which match the desired solutions. 5. Therefore, the quadratic function with solutions at $1$ and $-\frac{3}{4}$ is option D: $(x - 1)(4x + 3) = 0$. Final answer: Option D