The table below shows the values of the relation \(y = 11 - 2x - 2x^{2}\) for \(-4 \leq x \leq 3\). x -4 -3 -2 -1 0 1 2 3 y -13 11 (a) Copy and complete the...

(a) Copy and complete the table

Using \(y = 11 - 2x - 2x^{2}\), build the value from its parts \(x^{2}\), \(11\), \(-2x\) and \(-2x^{2}\) for each \(x\):

\(x=-3:\ 11-2(-3)-2(9)=11+6-18=-1\)
\(x=-2:\ 11+4-8=7\)
\(x=-1:\ 11+2-2=11\)
\(x=1:\ 11-2-2=7\)
\(x=2:\ 11-4-8=-1\)
\(x=3:\ 11-6-18=-13\)

(b) The graph of \(y = 11 - 2x - 2x^{2}\)

Plotting the eight points \((-4,-13),(-3,-1),(-2,7),(-1,11),(0,11),(1,7),(2,-1),(3,-13)\) with a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, and joining them with a smooth curve, gives a downward (maximum) parabola with its peak midway between \(x=-1\) and \(x=0\), at about \((-0.5,\ 11.5)\).

(c) Using the graph

(i) Roots of \(11 - 2x - 2x^{2} = 0\). These are the \(x\)-values where the curve crosses the \(x\)-axis (\(y=0\)). Reading off the graph:

(ii) Values of \(x\) for which \(3 - 2x - 2x^{2} = 0\). Subtracting this equation from the curve:

So where \(3 - 2x - 2x^{2}=0\) the curve has \(y = 8\). Drawing the horizontal line \(y = 8\) (shown dashed) and reading where it meets the curve:

(iii) Gradient of the curve at \(x = 1\). Draw the tangent to the curve at the point \((1,\ 7)\) (shown as the straight line touching the curve). Taking two points on this tangent, e.g. \((0,\ 13)\) and \((2,\ 1)\):

Hence the gradient of the curve at \(x = 1\) is \(-6\).