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

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

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

Answer

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.)