1. **Problem Statement:** Given the function $f(x) = -x^2 - 3x + 28$, find the x- and y-intercepts, the vertex coordinates and its nature, plot the function, and find the axis of symmetry. 2. **Finding the intercepts:** - The y-intercept is found by evaluating $f(0)$: $$f(0) = -0^2 - 3\times0 + 28 = 28$$ So, the y-intercept is at $(0, 28)$. - The x-intercepts are found by solving $f(x) = 0$: $$-x^2 - 3x + 28 = 0$$ Multiply both sides by $-1$ to simplify: $$x^2 + 3x - 28 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=3$, $c=-28$: $$x = \frac{-3 \pm \sqrt{3^2 - 4\times1\times(-28)}}{2\times1} = \frac{-3 \pm \sqrt{9 + 112}}{2} = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ So, $$x_1 = \frac{-3 + 11}{2} = 4, \quad x_2 = \frac{-3 - 11}{2} = -7$$ The x-intercepts are at $(4, 0)$ and $(-7, 0)$. 3. **Finding the vertex:** - The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$ where $a = -1$, $b = -3$: $$x = -\frac{-3}{2 \times -1} = -\frac{-3}{-2} = -\frac{3}{2} = -1.5$$ - Substitute $x = -1.5$ into $f(x)$ to find the y-coordinate: $$f(-1.5) = -(-1.5)^2 - 3(-1.5) + 28 = -2.25 + 4.5 + 28 = 30.25$$ - The vertex is at $(-1.5, 30.25)$. - Since $a = -1 < 0$, the parabola opens downward, so the vertex is a maximum point. 4. **Axis of symmetry:** - The axis of symmetry is the vertical line passing through the vertex: $$x = -1.5$$ 5. **Summary:** - X-intercepts: $(4, 0)$ and $(-7, 0)$ - Y-intercept: $(0, 28)$ - Vertex: $(-1.5, 30.25)$ (maximum point) - Axis of symmetry: $x = -1.5$