11. Sketch the function: $$ f(x)=-\frac{1}{4}(x+5)(x-1)^{2} $$. Label $$ x $$ and $$ y $$ intercepts. Compute, plot and label the points $$ (-6, f(-6)),(-5, f(-5)),(-4, f(-4)),(-3, f(-3)),(-2, f(-2)),(-1, f(-1)),(0, f(0)) $$, $$ (1, f(1)),(2, f(2)) $$. Graph MUST show the effect of any multiplicities greater than 1 .

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Alright, I need to sketch the function $$ f(x) = -\frac{1}{4}(x + 5)(x - 1)^2 $$. First, let's identify the key features of this function to understand its behavior.

**1. Finding the x-intercepts:**
The x-intercepts occur where $$ f(x) = 0 $$. Setting the function equal to zero:
$$
-\frac{1}{4}(x + 5)(x - 1)^2 = 0
$$
This equation is satisfied when either $$ x + 5 = 0 $$ or $$ (x - 1)^2 = 0 $$.
- $$ x + 5 = 0 $$ gives $$ x = -5 $$.
- $$ (x - 1)^2 = 0 $$ gives $$ x = 1 $$.

So, the x-intercepts are at $$ x = -5 $$ and $$ x = 1 $$.

**2. Finding the y-intercept:**
The y-intercept occurs when $$ x = 0 $$:
$$
f(0) = -\frac{1}{4}(0 + 5)(0 - 1)^2 = -\frac{1}{4} \times 5 \times 1 = -\frac{5}{4}
$$
So, the y-intercept is at $$ y = -\frac{5}{4} $$.

**3. Determining the behavior around the x-intercepts:**
- At $$ x = -5 $$, the factor $$ (x + 5) $$ has a multiplicity of 1. This means the graph crosses the x-axis at this point.
- At $$ x = 1 $$, the factor $$ (x - 1)^2 $$ has a multiplicity of 2. This indicates that the graph touches the x-axis and turns around, creating a local minimum or maximum at this point.

**4. Plotting the given points:**
Let's compute the function values for the specified x-values:
- $$ f(-6) = -\frac{1}{4}(-6 + 5)(-6 - 1)^2 = -\frac{1}{4}(-1)(49) = \frac{49}{4} = 12.25 $$
- $$ f(-5) = 0 $$ (as established)
- $$ f(-4) = -\frac{1}{4}(-4 + 5)(-4 - 1)^2 = -\frac{1}{4}(1)(25) = -\frac{25}{4} = -6.25 $$
- $$ f(-3) = -\frac{1}{4}(-3 + 5)(-3 - 1)^2 = -\frac{1}{4}(2)(16) = -\frac{32}{4} = -8 $$
- $$ f(-2) = -\frac{1}{4}(-2 + 5)(-2 - 1)^2 = -\frac{1}{4}(3)(9) = -\frac{27}{4} = -6.75 $$
- $$ f(-1) = -\frac{1}{4}(-1 + 5)(-1 - 1)^2 = -\frac{1}{4}(4)(4) = -\frac{16}{4} = -4 $$
- $$ f(0) = -\frac{5}{4} $$ (as established)
- $$ f(1) = 0 $$ (as established)
- $$ f(2) = -\frac{1}{4}(2 + 5)(2 - 1)^2 = -\frac{1}{4}(7)(1) = -\frac{7}{4} = -1.75 $$

**5. Plotting the Points:**
Plotting these points on a graph:
- $$ (-6, 12.25) $$
- $$ (-5, 0) $$
- $$ (-4, -6.25) $$
- $$ (-3, -8) $$
- $$ (-2, -6.75) $$
- $$ (-1, -4) $$
- $$ (0, -1.25) $$
- $$ (1, 0) $$
- $$ (2, -1.75) $$

**6. Graphing the Function:**
To graph $$ f(x) = -\frac{1}{4}(x + 5)(x - 1)^2 $$, consider the following:
- The leading term is $$ -\frac{1}{4}x^3 $$, indicating that as $$ x $$ approaches positive infinity, $$ f(x) $$ approaches negative infinity, and as $$ x $$ approaches negative infinity, $$ f(x) $$ approaches positive infinity.
- The function has a local maximum at $$ x = 1 $$ due to the multiplicity of 2 in $$ (x - 1)^2 $$.
- The graph crosses the x-axis at $$ x = -5 $$ and touches it at $$ x = 1 $$.

**7. Final Graph:**
The graph will show:
- A cubic-like shape with the specified intercepts.
- The point $$ (-5, 0) $$ where the graph crosses the x-axis.
- The point $$ (1, 0) $$ where the graph touches the x-axis and changes direction.
- The y-intercept at $$ (0, -1.25) $$.
- The points $$ (-6, 12.25) $$, $$ (-4, -6.25) $$, $$ (-3, -8) $$, $$ (-2, -6.75) $$, $$ (-1, -4) $$, $$ (2, -1.75) $$ plotted accordingly.

This sketch should accurately represent the function's behavior, highlighting the effects of the multiplicities on the graph.

**Graph of $$ f(x) = -\frac{1}{4}(x + 5)(x - 1)^2 $$:** - **x-intercepts:** $$ x = -5 $$ and $$ x = 1 $$ - **y-intercept:** $$ y = -\frac{5}{4} $$ at $$ x = 0 $$ **Points Plotted:** - $$ (-6, 12.25) $$ - $$ (-5, 0) $$ - $$ (-4, -6.25) $$ - $$ (-3, -8) $$ - $$ (-2, -6.75) $$ - $$ (-1, -4) $$ - $$ (0, -1.25) $$ - $$ (1, 0) $$ - $$ (2, -1.75) $$ The graph shows the function crossing the x-axis at $$ x = -5 $$ and touching it at $$ x = 1 $$ due to the multiplicity of 2 in $$ (x - 1)^2 $$. As $$ x $$ approaches positive infinity, $$ f(x) $$ approaches negative infinity, and as $$ x $$ approaches negative infinity, $$ f(x) $$ approaches positive infinity. The local maximum occurs at $$ x = 1 $$.

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