1. **Problem Statement:** Given the function $f(x) = -x^2 - 3x + 28$, find the X and Y intercepts, compute the vertex coordinates and its nature, plot the function, and draw the axis of symmetry with its equation. 2. **Formula and Rules:** - The general quadratic form is $ax^2 + bx + c$. - X-intercepts are found by solving $f(x) = 0$. - Y-intercept is $f(0) = c$. - Vertex coordinates are given by $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$. - The axis of symmetry is the vertical line $x = -\frac{b}{2a}$. - Since $a < 0$, the parabola opens downward, so the vertex is a maximum point. 3. **Find X-intercepts:** Solve $-x^2 - 3x + 28 = 0$. Multiply both sides by $-1$ for simplicity: $x^2 + 3x - 28 = 0$. Use quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=3$, $c=-28$. Calculate discriminant: $\Delta = 3^2 - 4(1)(-28) = 9 + 112 = 121$. Square root: $\sqrt{121} = 11$. So, $x = \frac{-3 \pm 11}{2}$. Two solutions: - $x = \frac{-3 + 11}{2} = \frac{8}{2} = 4$ - $x = \frac{-3 - 11}{2} = \frac{-14}{2} = -7$ 4. **Find Y-intercept:** Evaluate $f(0) = -0^2 - 3(0) + 28 = 28$. 5. **Find Vertex Coordinates:** Calculate $x$-coordinate of vertex: $x_v = -\frac{b}{2a} = -\frac{-3}{2(-1)} = -\frac{-3}{-2} = -\frac{3}{2} = -1.5$. Calculate $y$-coordinate: $f(-1.5) = -(-1.5)^2 - 3(-1.5) + 28 = -2.25 + 4.5 + 28 = 30.25$. Vertex is at $(-1.5, 30.25)$. Since $a = -1 < 0$, vertex is a maximum point. 6. **Axis of Symmetry:** Equation is $x = -1.5$. 7. **Summary:** - X-intercepts: $(4, 0)$ and $(-7, 0)$ - Y-intercept: $(0, 28)$ - Vertex: $(-1.5, 30.25)$ (maximum) - Axis of symmetry: $x = -1.5$