Algebra, Year 11: perpendicular lines, roots and turning points

−1

HintTry it with the gradients 2 and −1/2.

The whyFor example y = 2x + 1 and y = −(1/2)x + 4 meet at right angles, and 2 × (−1/2) = −1. So each gradient is the negative reciprocal of the other.

−3/2

HintTurn the gradient upside down and change its sign.

The whyThe given line has gradient 2/3. The perpendicular gradient is the negative reciprocal, −3/2. Check: (2/3) × (−3/2) = −1.

y = −(1/2)x + 3

HintFind the new gradient first, then use the point to find the intercept.

The whyThe perpendicular gradient is −1/2, so y = −(1/2)x + c. Substituting (4, 1): 1 = −2 + c, so c = 3. Both lines happen to cross the y-axis at 3.

Yes: their gradients are 3 and −1/3, and 3 × (−1/3) = −1

HintRearrange the second equation into the form y = mx + c.

The whyx + 3y = 6 rearranges to y = −(1/3)x + 2, so its gradient is −1/3. The product of the gradients is −1, which is the test for lines at right angles.

The negative reciprocal of a number

HintA two-word name: the first word describes the sign change, the second the flipping.

The whyIf a line has gradient m, every line perpendicular to it has gradient −1/m. For a whole number such as 4, write it as 4/1 first, giving −1/4.

x = −2 or x = −3

HintFactorise the left-hand side into two brackets first.

The whyx² + 5x + 6 = (x + 2)(x + 3). A product is zero only if one of its factors is zero, so x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.

(x − 2)(x − 5) = 0

HintEach solution must make one of the brackets equal to zero.

The whyx = 2 makes (x − 2) zero and x = 5 makes (x − 5) zero, so (x − 2)(x − 5) = 0. Expanded, this is x² − 7x + 10 = 0. Working backwards like this shows why the signs in the brackets are the opposite of the solutions.

x = 0 or x = 7

HintThere is no number term, so take out a common factor.

The whyx² − 7x = x(x − 7). So x = 0 or x − 7 = 0, which gives x = 0 or x = 7. Zero is a perfectly good solution and must be written down.

(−2, 0) and (4, 0)

HintOn the x-axis, y is zero.

The whySet y = 0: x² − 2x − 8 = 0, which factorises to (x + 2)(x − 4) = 0, so x = −2 or x = 4. These roots are the x-coordinates of the crossing points.

3

HintFactorise, and notice that both brackets are the same.

The whyx² − 6x + 9 = (x − 3)(x − 3) = (x − 3)². Both brackets give the same answer, x = 3, called a repeated root. On a graph, the curve just touches the x-axis there.

x = 5 or x = −2

HintCollect every term on one side so that the other side is zero.

The whyRearranged, x² − 3x − 10 = 0, which factorises to (x − 5)(x + 2) = 0, so x = 5 or x = −2. Factorising only helps when one side is zero, because only a product equal to zero forces one factor to be zero.

(3, 2)

HintWhich value of x makes the squared bracket as small as it can be?

The why(x − 3)² is zero when x = 3 and positive everywhere else, so the lowest point is at x = 3, where y = 0 + 2 = 2. In y = (x + p)² + q the turning point is (−p, q).

(−3, −4)

HintRewrite the right-hand side as a squared bracket plus or minus a number.

The whyx² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. The bracket is zero when x = −3, and then y = −4, so the minimum point is (−3, −4).

5

HintThe line of symmetry passes through the turning point.

The whyx² − 10x + 3 = (x − 5)² − 25 + 3 = (x − 5)² − 22. The turning point is (5, −22), and a quadratic graph is symmetrical about the vertical line through its turning point, x = 5.

(x − 3)² is never negative: its smallest value is 0, when x = 3, and adding 2 gives 2

HintWhat can you say about the sign of a square?

The whyA square is zero or positive, so (x − 3)² ≥ 0 for every x. The whole expression is therefore at least 2, and equals 2 only at x = 3. This is why completed-square form shows the turning point at a glance.

y = (x − 2)² − 9

HintThe bracket must be zero at the turning point's x-value.

The whyThe turning point (2, −9) means the bracket is zero when x = 2, so it is (x − 2), and the number added is −9. Expanding gives y = x² − 4x − 5, so b = −4 and c = −5.