Question

graph the function.
$f(x) = -dfrac{1}{5}x^2 + 2x - 4$

Explanation:

Step1: Identify the vertex form of the parabola

The given function is \( f(x) = -\frac{1}{5}x^2 + 2x - 4 \). For a quadratic function \( ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{5} \) and \( b = 2 \).
\( x = -\frac{2}{2\times(-\frac{1}{5})}=-\frac{2}{-\frac{2}{5}} = 5 \)

Step2: Find the y - coordinate of the vertex

Substitute \( x = 5 \) into the function \( f(x) \):
\( f(5)=-\frac{1}{5}(5)^2+2(5)-4=-\frac{1}{5}\times25 + 10-4=- 5 + 10 - 4=1 \)? Wait, but the graph shows the vertex at (5,9). Wait, maybe we made a mistake. Let's recalculate:
\( f(5)=-\frac{1}{5}(25)+2\times5 - 4=-5 + 10-4 = 1 \). Wait, the graph has a vertex at (5,9). Maybe the function is \( f(x)=-\frac{1}{5}x^{2}+2x + 4 \)? No, the given function is \( f(x)=-\frac{1}{5}x^{2}+2x - 4 \). Wait, maybe the graph is for a different function, but we can analyze the given function.
The y - intercept: when \( x = 0 \), \( f(0)=-4 \), which matches the green dot at (0, - 4).
To graph the function:

• The parabola opens downwards because \( a=-\frac{1}{5}<0 \).
• Vertex at \( (h,k)=(5,1) \) (from calculation), but the graph shows vertex at (5,9). Maybe there is a typo, but assuming the function is as given:

We can find two more points. Let's take \( x = 1 \): \( f(1)=-\frac{1}{5}(1)+2(1)-4=-\frac{1}{5}+2 - 4=-\frac{1}{5}-2=-\frac{11}{5}=-2.2 \)
\( x = 10 \): \( f(10)=-\frac{1}{5}(100)+20 - 4=-20 + 20-4=-4 \)
So the points are (0, - 4), (5,1), (10, - 4), (1, - 2.2) etc. We can plot these points and draw a parabola opening downwards.

(Note: There seems to be a discrepancy between the calculated vertex and the graph's vertex. Maybe the function in the graph is \( f(x)=-\frac{1}{5}x^{2}+2x + 4 \), in which case \( f(5)=-\frac{1}{5}(25)+10 + 4=-5 + 10 + 4 = 9 \), which matches the graph's vertex at (5,9). If we assume the function is \( f(x)=-\frac{1}{5}x^{2}+2x + 4 \)):

Step1: Find x - coordinate of vertex

\( a = -\frac{1}{5},b = 2 \), \( x=-\frac{b}{2a}=-\frac{2}{2\times(-\frac{1}{5})}=5 \)

Step2: Find y - coordinate of vertex

\( f(5)=-\frac{1}{5}(25)+2(5)+4=-5 + 10 + 4 = 9 \)

Step3: Y - intercept

When \( x = 0 \), \( f(0)=4 \)? No, the graph has y - intercept at (0, - 4). So there is a confusion. But following the given function \( f(x)=-\frac{1}{5}x^{2}+2x - 4 \):

To graph:

1. Plot the y - intercept (0, - 4).
2. Plot the vertex (5,1).
3. Plot the point symmetric to the y - intercept with respect to the vertex. The axis of symmetry is \( x = 5 \), the distance from \( x = 0 \) to \( x = 5 \) is 5 units, so the symmetric point is \( x=5 + 5=10 \), \( f(10)=-4 \), so (10, - 4).
4. Draw a parabola opening downwards through these points.

If we consider the graph shown, with vertex at (5,9) and y - intercept at (0, - 4), let's find the function. Let the vertex form be \( f(x)=a(x - 5)^2+9 \). Substitute (0, - 4):
\( -4=a(0 - 5)^2+9\Rightarrow -4 = 25a+9\Rightarrow 25a=-13\Rightarrow a=-\frac{13}{25}=-0.52 \), but the given function has \( a = -\frac{1}{5}=-0.2 \). So there is a mismatch, but we proceed with the given function.

Answer:

To graph \( f(x)=-\frac{1}{5}x^{2}+2x - 4 \):

• Direction: Opens downward (since \( a = -\frac{1}{5}<0 \)).
• Vertex: \( (5,1) \) (calculated as \( x = 5,y = f(5)=1 \)).
• Y - intercept: \( (0, - 4) \).
• Symmetric point to y - intercept: \( (10, - 4) \) (since axis of symmetry is \( x = 5 \), distance from \( x = 0 \) to \( x = 5 \) is 5, so \( x=5 + 5 = 10 \)).
• Plot these points and draw a smooth parabola through them.

(If the graph's vertex is correct, the function might have a typo, but based on the given function, the above steps are for graphing it.)