(a) Copy and complete the table of values for the relation \(y = -x^{2} + x + 2; -3 \leq x \leq 3\). x -3 -2 -1 0 1 2 3 y -4 2 -4 (b) Using scales of 2 cm t...

Answer Details

(a) Completing the table of values for \(y=-x^{2}+x+2\).

Evaluate the relation at each missing value of \(x\):

\(x=-3:\ y=-(-3)^{2}+(-3)+2=-9-3+2=-10\)

\(x=-1:\ y=-(-1)^{2}+(-1)+2=-1-1+2=0\)

\(x=1:\ y=-(1)^{2}+(1)+2=-1+1+2=2\)

\(x=2:\ y=-(2)^{2}+(2)+2=-4+2+2=0\)

The completed table is:

(b) Graph of \(y=-x^{2}+x+2\).

Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 2 units on the \(y\)-axis, the seven points \((-3,-10),(-2,-4),(-1,0),(0,2),(1,2),(2,0),(3,-4)\) are plotted and joined with a smooth curve. Because the coefficient of \(x^{2}\) is negative, the curve opens downward with its turning point (a maximum) near \(x=0.5\).

(c) Readings from the graph.

(i) Turning value of \(y\). The curve turns at \(x=\tfrac{1}{2}\), where

\[y=-(0.5)^{2}+0.5+2=-0.25+0.5+2=2.25.\]

Since the parabola opens downward, this turning point is a maximum, so the turning value read from the graph is \(y=2.25\) at \(x=0.5\). Over the drawn range \(-3\le x\le 3\) the least (lowest) value reached is \(y=-10\) at \(x=-3\).

(ii) Roots of \(x^{2}-x-2=0\). Since \(-x^{2}+x+2=-(x^{2}-x-2)\), the equation \(x^{2}-x-2=0\) is equivalent to \(y=0\). The curve cuts the \(x\)-axis where \(y=0\), and from the graph these crossings are at

\[x=-1\quad\text{and}\quad x=2.\]

(iii) Gradient of the curve at \(x=-0.5\). A tangent is drawn to the curve at the point \((-0.5,\,1.25)\). Taking a convenient right-angled triangle along the tangent, the change in \(y\) divided by the change in \(x\) gives

\[\text{gradient}=\frac{\Delta y}{\Delta x}\approx 2.\]

(This agrees with the exact slope: \(\dfrac{dy}{dx}=-2x+1\), so at \(x=-0.5\), \(\dfrac{dy}{dx}=-2(-0.5)+1=1+1=2\).) The gradient of the curve at \(x=-0.5\) is therefore approximately \(2\).