Question

graph the function ( f(x) = -\frac{1}{3}x^2 - \frac{2}{3}x + \frac{5}{3} ).
plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Find the x-coordinate of the vertex

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

Step2: Find the y-coordinate of the vertex

Substitute \( x = -1 \) into the function \( f(x) = -\frac{1}{3}x^2 - \frac{2}{3}x + \frac{5}{3} \). So, \( f(-1) = -\frac{1}{3}(-1)^2 - \frac{2}{3}(-1) + \frac{5}{3} = -\frac{1}{3} + \frac{2}{3} + \frac{5}{3} = \frac{-1 + 2 + 5}{3} = \frac{6}{3} = 2 \). So the vertex is at \( (-1, 2) \).

Step3: Find another point on the parabola

Let's choose \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = -\frac{1}{3}(0)^2 - \frac{2}{3}(0) + \frac{5}{3} = \frac{5}{3} \approx 1.67 \)? Wait, no, wait: \( f(0) = \frac{5}{3} \)? Wait, no, let's recalculate. Wait, \( f(0) = -\frac{1}{3}(0) - \frac{2}{3}(0) + \frac{5}{3} = \frac{5}{3} \approx 1.67 \)? But maybe a better point. Let's choose \( x = 2 \). Then \( f(2) = -\frac{1}{3}(4) - \frac{2}{3}(2) + \frac{5}{3} = -\frac{4}{3} - \frac{4}{3} + \frac{5}{3} = \frac{-4 -4 +5}{3} = \frac{-3}{3} = -1 \). So the point \( (2, -1) \) is on the parabola. Or \( x = 1 \): \( f(1) = -\frac{1}{3}(1) - \frac{2}{3}(1) + \frac{5}{3} = \frac{-1 -2 +5}{3} = \frac{2}{3} \approx 0.67 \). Alternatively, \( x = -2 \): \( f(-2) = -\frac{1}{3}(4) - \frac{2}{3}(-2) + \frac{5}{3} = -\frac{4}{3} + \frac{4}{3} + \frac{5}{3} = \frac{5}{3} \approx 1.67 \). But let's take \( x = 1 \), \( f(1) = \frac{2}{3} \), or \( x = 0 \), \( f(0) = \frac{5}{3} \). Wait, maybe a simpler point. Let's check \( x = 1 \): \( f(1) = -\frac{1}{3}(1) - \frac{2}{3}(1) + \frac{5}{3} = \frac{-1 -2 +5}{3} = \frac{2}{3} \). Or \( x = 2 \), \( f(2) = -1 \). So the vertex is at \( (-1, 2) \), and another point could be \( (0, \frac{5}{3}) \) or \( (2, -1) \) or \( (1, \frac{2}{3}) \).

Answer:

The vertex of the parabola is at \((-1, 2)\). Another point on the parabola can be \((0, \frac{5}{3})\) (or other valid points like \((2, -1)\), \((1, \frac{2}{3})\) etc.). To graph, plot the vertex \((-1, 2)\) and then plot the chosen point (e.g., \((0, \frac{5}{3})\)) and draw the parabola opening downward (since \(a = -\frac{1}{3} < 0\)).