1. **State the problem:** Find the vertex, y-intercept, and x-intercepts of the quadratic function $$y = 4x^2 - 8x - 5$$. 2. **Rewrite the quadratic in vertex form:** Start by factoring out 4 from the first two terms: $$y = 4(x^2 - 2x) - 5$$ 3. **Complete the square inside the parentheses:** Take half of the coefficient of $x$, which is $-2$, half is $-1$, square it to get $1$. Add and subtract $1$ inside the parentheses: $$y = 4(x^2 - 2x + 1 - 1) - 5 = 4((x - 1)^2 - 1) - 5$$ 4. **Simplify:** $$y = 4(x - 1)^2 - 4 - 5 = 4(x - 1)^2 - 9$$ 5. **Identify the vertex:** The vertex form is $$y = a(x - h)^2 + k$$ where vertex is $(h, k)$. Here, vertex is at $$ (1, -9) $$. 6. **Find the y-intercept:** Set $$x = 0$$: $$y = 4(0)^2 - 8(0) - 5 = -5$$ So y-intercept is $$ (0, -5) $$. 7. **Find the x-intercepts:** Set $$y = 0$$: $$0 = 4x^2 - 8x - 5$$ Divide both sides by 4: $$0 = \cancel{4}x^2 - \cancel{4}2x - \cancel{4}\frac{5}{4}$$ $$0 = x^2 - 2x - \frac{5}{4}$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1$$, $$b=-2$$, $$c=-\frac{5}{4}$$. Calculate discriminant: $$b^2 - 4ac = (-2)^2 - 4(1)(-\frac{5}{4}) = 4 + 5 = 9$$ Calculate roots: $$x = \frac{2 \pm \sqrt{9}}{2} = \frac{2 \pm 3}{2}$$ So, $$x_1 = \frac{2 + 3}{2} = \frac{5}{2} = 2.5$$ $$x_2 = \frac{2 - 3}{2} = \frac{-1}{2} = -0.5$$ Thus, x-intercepts are $$\left(2.5, 0\right)$$ and $$\left(-0.5, 0\right)$$. **Final answers:** - Vertex: $$(1, -9)$$ - Y-intercept: $$(0, -5)$$ - X-intercepts: $$(2.5, 0)$$ and $$(-0.5, 0)$$