Question

graph the equation.
$y = -dfrac{1}{8}x^2 + x - 4$

Explanation:

Step1: Identify the parabola's direction and vertex

The equation is \( y = -\frac{1}{8}x^2 + x - 4 \). The coefficient of \( x^2 \) is negative (\( -\frac{1}{8} \)), so the parabola opens downward. Wait, but the given graph opens upward. Wait, maybe I made a mistake. Wait, let's recalculate the vertex. The vertex of a parabola \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{8} \), \( b = 1 \). So \( x = -\frac{1}{2 \times (-\frac{1}{8})} = -\frac{1}{-\frac{1}{4}} = 4 \). Then \( y = -\frac{1}{8}(4)^2 + 4 - 4 = -\frac{1}{8} \times 16 + 0 = -2 \). So the vertex is at (4, -2). But the given graph has a vertex at (0,0) and opens upward. Wait, maybe the graph is incorrect, or maybe I misread the equation. Wait, the equation is \( y = -\frac{1}{8}x^2 + x - 4 \)? Wait, no, maybe it's \( y = \frac{1}{8}x^2 - x + 4 \)? No, the original equation is \( y = -\frac{1}{8}x^2 + x - 4 \). Wait, let's check the y-intercept. When \( x = 0 \), \( y = -4 \). But the graph has a y-intercept at (0,0). So there's a discrepancy. Wait, maybe the equation is \( y = \frac{1}{8}x^2 - x \)? No, the problem says \( y = -\frac{1}{8}x^2 + x - 4 \). Wait, perhaps the graph is not the correct one for this equation. But maybe the task is to graph the equation correctly. Let's proceed.

Step2: Rewrite the equation in vertex form

Complete the square for \( y = -\frac{1}{8}x^2 + x - 4 \). Factor out \( -\frac{1}{8} \) from the first two terms: \( y = -\frac{1}{8}(x^2 - 8x) - 4 \). Now, complete the square inside the parentheses: \( x^2 - 8x \) needs \( (\frac{-8}{2})^2 = 16 \). So add and subtract 16 inside the parentheses: \( y = -\frac{1}{8}(x^2 - 8x + 16 - 16) - 4 = -\frac{1}{8}((x - 4)^2 - 16) - 4 = -\frac{1}{8}(x - 4)^2 + 2 - 4 = -\frac{1}{8}(x - 4)^2 - 2 \). So the vertex is at (4, -2), and it opens downward.

Step3: Plot key points

• Vertex: (4, -2)
• Y-intercept: when \( x = 0 \), \( y = -4 \), so (0, -4)
• X-intercepts: set \( y = 0 \), \( -\frac{1}{8}x^2 + x - 4 = 0 \). Multiply both sides by -8: \( x^2 - 8x + 32 = 0 \). Discriminant: \( (-8)^2 - 4 \times 1 \times 32 = 64 - 128 = -64 < 0 \), so no real x-intercepts.

So the parabola opens downward, vertex at (4, -2), y-intercept at (0, -4), and no x-intercepts. The given graph is incorrect (opens upward, vertex at (0,0), y-intercept at (0,0)), so we need to graph the correct parabola.

Answer:

To graph \( y = -\frac{1}{8}x^2 + x - 4 \):

1. Vertex: \( (4, -2) \) (from \( x = -\frac{b}{2a} = 4 \), \( y = -\frac{1}{8}(4)^2 + 4 - 4 = -2 \)).
2. Direction: Opens downward (since \( a = -\frac{1}{8} < 0 \)).
3. Key Points:

• Y-intercept: \( (0, -4) \) (substitute \( x = 0 \)).
• No real x-intercepts (discriminant \( < 0 \)).

Plot the vertex, y-intercept, and sketch the downward-opening parabola. The given graph in the problem is incorrect for this equation.