1. **Problem Statement:** Given the function $f(x) = -x^2 - 3x + 28$, we need to find: a) The $x$ and $y$ intercepts. b) The coordinates of the vertex and its nature. c) Plot the function. d) Draw the axis of symmetry and find its equation. 2. **Finding the intercepts:** - The $y$-intercept is found by evaluating $f(0)$: $$f(0) = -(0)^2 - 3(0) + 28 = 28$$ 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 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=3$, $c=-28$: $$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-28)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 112}}{2} = \frac{-3 \pm \sqrt{121}}{2}$$ $$x = \frac{-3 \pm 11}{2}$$ So, $$x_1 = \frac{-3 + 11}{2} = 4$$ $$x_2 = \frac{-3 - 11}{2} = -7$$ The $x$-intercepts are at $(4, 0)$ and $(-7, 0)$. 3. **Finding the vertex:** The vertex of a parabola $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Here, $a = -1$, $b = -3$. $$x_v = -\frac{-3}{2(-1)} = \frac{3}{-2} = -1.5$$ Find $y$ coordinate by evaluating $f(-1.5)$: $$f(-1.5) = -(-1.5)^2 - 3(-1.5) + 28 = -2.25 + 4.5 + 28 = 30.25$$ So, 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. **Plotting the function:** The function is a downward opening parabola with vertex at $(-1.5, 30.25)$, $x$-intercepts at $(4,0)$ and $(-7,0)$, and $y$-intercept at $(0,28)$. Final answers: - $x$-intercepts: $(4, 0)$ and $(-7, 0)$ - $y$-intercept: $(0, 28)$ - Vertex: $(-1.5, 30.25)$ (maximum point) - Axis of symmetry: $x = -1.5$