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 .

To sketch the function \( f(x) = -\frac{1}{4}(x+5)(x-1)^{2} \), start by identifying the x-intercepts and y-intercept. The x-intercepts occur when \( f(x)=0 \). Setting the equation equal to zero gives us: 1. From \( x + 5 = 0 \), we get \( x = -5 \). 2. From \( (x - 1)^{2} = 0 \), we get \( x = 1 \) with a multiplicity of 2. The y-intercept occurs when \( x = 0 \): \[ f(0) = -\frac{1}{4}(0 + 5)(0 - 1)^{2} = -\frac{1}{4}(5)(1) = -\frac{5}{4}. \] Now, to compute the values: - \( f(-6) = -\frac{1}{4}(-6 + 5)(-6 - 1)^{2} = -\frac{1}{4}(-1)(49) = \frac{49}{4} \) - \( f(-5) = 0 \) - \( f(-4) = -\frac{1}{4}(1)(25) = -\frac{25}{4} \) - \( f(-3) = -\frac{1}{4}(2)(16) = -\frac{32}{4} = -8 \) - \( f(-2) = -\frac{1}{4}(3)(9) = -\frac{27}{4} \) - \( f(-1) = -\frac{1}{4}(4)(4) = -4 \) - \( f(0) = -\frac{5}{4} \) - \( f(1) = 0 \) - \( f(2) = -\frac{1}{4}(7)(1) = -\frac{7}{4} \) To sketch the graph: 1. Mark the intercepts at \( (-5, 0) \) and \( (1, 0) \) and note that at \( x = 1 \) the graph touches the x-axis but does not cross it due to the multiplicity of 2. 2. Plot the calculated points, showing a downward curvature because of the leading negative coefficient, and the vertex near these x-intercepts. 3. The graph will exhibit a parabola-like shape, decreasing before \( x = -5 \), coming back up to touch the x-axis at \( x = -5 \), and then further dipping down towards \( x = 1 \) and continuing downwards. Label your axes, and highlight the points you computed for clarity!