Higher June 2019 Paper 1 Q6
6. The table shows some of the values of \(y = x^2 + x - 1\) for \(-2 \leqslant x \leqslant 1\).
| \(x\) | \(-2\) | \(-1\) | \(-0{\cdot}5\) | \(0\) | \(1\) |
|---|---|---|---|---|---|
| \(y = x^2 + x - 1\) | \(-1\) | \(-1{\cdot}25\) |
(a) Complete the table above. [2]
(b) On the graph paper below, draw the graph of \(y = x^2 + x - 1\) for \(-2 \leqslant x \leqslant 1\). [2]
(c) State the equation of the line of symmetry of the curve \(y = x^2 + x - 1\). [1]
(d) Use your graph to solve \(x^2 + x - 1 = 0\). [2]
| Scheme | Marks | Comments | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B2 | B1 for any two correct |
| Scheme | Marks | Comments |
|---|---|---|
| All 5 correct points plotted correctly and joined with a smooth curve | B2 | Mark intent B1 for a smooth curve at least through 3 correct pairs of coordinates or for all of their 5 pairs of coordinates plotted correctly Allow 2 marks here if curve correct even if there is a slip in their table |
| Scheme | Marks | Comments |
|---|---|---|
| \(x = -\dfrac{1}{2}\) oe | B1 | Equation must be stated; check graph; not for \(x = -\dfrac{1}{2}\ \ y = -1.25\) |
| Scheme | Marks | Comments |
|---|---|---|
| (\(x\) =) 0.5 to 0.7 , \(-1.5\) to \(-1.7\) | B2 | or FT ‘their curve’ B1 for one correct root If their curve has more than 2 roots, they must give all their solutions for B2 and may omit one solution only for B1. |
| (7) |