Question

graph the function.
$h(x) = -dfrac{1}{3}x^2 + 2x - 4$

Explanation:

Step1: Identify the function type

The function \( h(x) = -\frac{1}{3}x^2 + 2x - 4 \) is a quadratic function (since the highest power of \( x \) is 2), so its graph should be a parabola. The coefficient of \( x^2 \) is \( -\frac{1}{3} \), which is negative, so the parabola should open downward. Wait, but the given graph in the image opens upward. Wait, maybe I made a mistake. Wait, let's recalculate the vertex and other properties.

First, for a quadratic function \( ax^2 + bx + c \), the vertex is at \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{3} \), \( b = 2 \). So \( x = -\frac{2}{2 \times (-\frac{1}{3})} = -\frac{2}{-\frac{2}{3}} = 3 \). Then \( h(3) = -\frac{1}{3}(3)^2 + 2(3) - 4 = -\frac{1}{3}(9) + 6 - 4 = -3 + 6 - 4 = -1 \). So the vertex is at \( (3, -1) \). Wait, the given graph has a vertex at (0,0) which is different. Wait, maybe the graph in the image is not the correct graph for \( h(x) \). But the problem says "Graph the function" \( h(x) = -\frac{1}{3}x^2 + 2x - 4 \). Let's find the y-intercept: when \( x = 0 \), \( h(0) = -4 \). So the y-intercept is (0, -4). Let's find the x-intercepts by setting \( h(x) = 0 \): \( -\frac{1}{3}x^2 + 2x - 4 = 0 \). Multiply both sides by -3: \( x^2 - 6x + 12 = 0 \). The discriminant is \( (-6)^2 - 4(1)(12) = 36 - 48 = -12 \), which is negative, so there are no real x-intercepts. So the parabola opens downward (since \( a = -\frac{1}{3} < 0 \)) with vertex at (3, -1), y-intercept at (0, -4), and no x-intercepts.

But the given graph in the image opens upward, has vertex at (0,0), and x-intercept at (0,0). So maybe the graph is incorrect, or perhaps there's a mistake. Wait, maybe I misread the function. Let me check again: \( h(x) = -\frac{1}{3}x^2 + 2x - 4 \). Yes. So to graph this function, we can follow these steps:

Step2: Find the vertex

As calculated, \( x = -\frac{b}{2a} = -\frac{2}{2 \times (-\frac{1}{3})} = 3 \). Then \( h(3) = -1 \), so vertex is (3, -1).

Step3: Find the y-intercept

When \( x = 0 \), \( h(0) = -4 \), so point (0, -4).

Step4: Find another point

Let's take \( x = 6 \): \( h(6) = -\frac{1}{3}(36) + 12 - 4 = -12 + 12 - 4 = -4 \). So point (6, -4).

Step5: Plot the vertex, y-intercept, and (6, -4), then draw the parabola opening downward (since \( a < 0 \)).

But the given graph in the image is a parabola opening upward with vertex at (0,0), which does not match \( h(x) \). So perhaps the image is a distractor, and we need to graph the correct parabola.

Answer:

To graph \( h(x) = -\frac{1}{3}x^2 + 2x - 4 \):

1. Vertex: \( (3, -1) \) (from \( x = -\frac{b}{2a} = 3 \), \( h(3) = -1 \)).
2. Y-intercept: \( (0, -4) \) (set \( x = 0 \)).
3. Symmetric point: \( (6, -4) \) (since parabola is symmetric about \( x = 3 \), \( 6 - 3 = 3 - 0 \)).
4. Draw a downward - opening parabola (since \( a = -\frac{1}{3} < 0 \)) through these points.

(Note: The provided graph in the image does not match \( h(x) \); the correct graph has vertex \( (3, -1) \), y - intercept \( (0, -4) \), and no x - intercepts.)