Question

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

Explanation:

Step1: Identify the vertex form of the quadratic function

The given function is \( h(x) = -\frac{1}{3}x^2 + 2x - 4 \). For a quadratic function in the form \( ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{3} \) and \( b = 2 \).
So, \( x = -\frac{2}{2\times(-\frac{1}{3})}=-\frac{2}{-\frac{2}{3}} = 3 \)

Step2: Find the y - coordinate of the vertex

Substitute \( x = 3 \) into the function \( h(x) \):
\( h(3)=-\frac{1}{3}(3)^2+2(3)-4=-\frac{1}{3}\times9 + 6-4=- 3 + 6 - 4=-1 \)
So the vertex of the parabola is at \( (3,-1) \)

Step3: Determine the direction of the parabola

Since \( a=-\frac{1}{3}<0 \), the parabola opens downwards.

Step4: Find the y - intercept

To find the y - intercept, set \( x = 0 \):
\( h(0)=-\frac{1}{3}(0)^2+2(0)-4=-4 \), so the y - intercept is \( (0, - 4) \)

Step5: Find additional points (optional)

We can find another point, for example, when \( x = 6 \):
\( h(6)=-\frac{1}{3}(6)^2+2(6)-4=-\frac{1}{3}\times36 + 12-4=-12 + 12-4=-4 \)

Now, to graph the function:

• Plot the vertex \( (3,-1) \)
• Plot the y - intercept \( (0, - 4) \) and the point \( (6,-4) \) (since the parabola is symmetric about the vertical line \( x = 3 \), the point \( (6,-4) \) is the mirror image of \( (0,-4) \) with respect to \( x = 3 \))
• Draw a parabola opening downwards passing through these points.

Answer:

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

1. Vertex: \((3, - 1)\) (opens downward as \(a=-\frac{1}{3}<0\)).
2. Y - intercept: \((0, - 4)\), symmetric point: \((6, - 4)\) (due to symmetry about \(x = 3\)).
3. Plot these points and draw a downward - opening parabola through them.