1. **State the problem:** We are given the quadratic function $f(x) = -8x^2 + 10$ and want to understand its graph and key features. 2. **Formula and rules:** This is a quadratic function in the form $f(x) = ax^2 + bx + c$ where $a = -8$, $b = 0$, and $c = 10$. - Since $a < 0$, the parabola opens downward. - The vertex form of a parabola is $f(x) = a(x - h)^2 + k$ where $(h,k)$ is the vertex. 3. **Find the vertex:** - The vertex $x$-coordinate is given by $h = -\frac{b}{2a} = -\frac{0}{2(-8)} = 0$. - Substitute $x=0$ into $f(x)$ to find $k$: $k = f(0) = -8(0)^2 + 10 = 10$. - So the vertex is at $(0, 10)$. 4. **Find the y-intercept:** - The y-intercept is $f(0) = 10$ (already found). 5. **Find the x-intercepts:** - Set $f(x) = 0$: $$-8x^2 + 10 = 0$$ - Rearrange: $$-8x^2 = -10$$ $$\cancel{-8}x^2 = \cancel{-10}$$ - Divide both sides by $-8$: $$x^2 = \frac{10}{8} = \frac{5}{4}$$ - Take square root: $$x = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$$ 6. **Summary:** - The parabola opens downward. - Vertex at $(0, 10)$. - Y-intercept at $(0, 10)$. - X-intercepts at $\left(\frac{\sqrt{5}}{2}, 0\right)$ and $\left(-\frac{\sqrt{5}}{2}, 0\right)$. This describes the graph of $f(x) = -8x^2 + 10$ on the coordinate plane.