Cubic and Reciprocal Graphs

Try a negative value and a positive value of x in y=x^3. Where would those points be?

Can x be 0 in y=\frac{1}{x}? What does that mean for the graph and the y-axis?

x^3 means x\times x\times x, not 3\times x.

Cubing a negative number gives a negative answer.

\frac{12}{x} means 12 divided by x.

A positive number divided by a negative number gives a negative answer.

1. \frac{8}{x} means 8\div x.
2. 8\div(-2)=-4
3. y=-4

Which number can you never divide by?

1. \frac{1}{x} means 1\div x.
2. You cannot divide by 0.
3. So x cannot be 0. This is why the graph of y=\frac{1}{x} never meets the y-axis.

Try a positive value of x and a negative value of x. What sign does y have each time?

1. When x is positive, \frac{1}{x} is positive. For example, x=2 gives y=\frac{1}{2}.
2. When x is negative, \frac{1}{x} is negative. For example, x=-2 gives y=-\frac{1}{2}.
3. So one part has x and y both positive, and the other part has x and y both negative.

Each small square across is 0.5 and each small square up is 1. Find 1.5 on the x-axis first.

1. Go up from 1.5 on the x-axis to the curve, then across to the y-axis.
2. The curve is a little below y=2.5 there.
3. y is about 2.4. (Working it out: 1.5^3-1=3.375-1=2.375)

Can x be 0? Can \frac{1}{x} ever be 0? Think about what that means for the axes.

1. Kai's curves are in the right places: one where x and y are both positive, one where they are both negative.
2. But x can never be 0, and y can never be 0, so the graph of y=\frac{1}{x} never touches either axis.
3. Kai's curves meet the axes, so that is what is wrong.

Divide 4 by the value of x. Dividing by 0.5 gives a bigger answer.

Part (a)

1. y=4\div0.5
2. There are 8 halves in 4, so y=8.

Part (b)

1. y=4\div8=0.5

Every point on the y-axis has an x-coordinate of 0.

1. On the y-axis, x=0.
2. y=0^3-7=-7
3. The coordinates are (0, -7).

Which equations give straight lines and which give curves? Then try x = 2 in each equation: is y positive or negative?

Part (a)

1. y=-x^3 is a curve, not a straight line, so it is graph B or graph C.
2. When x=2, y=-2^3=-8, which is negative. So the curve falls from left to right.
3. This is graph C.

Part (b)

1. y=-x is a straight line through O, so it is graph A or graph D.
2. When x=2, y=-2, which is negative. So the line falls from left to right.
3. This is graph A.

Find 5 on the y-axis. It is halfway between 4 and 6.

1. Go across from 5 on the y-axis to the curve, then down to the x-axis.
2. The curve is at y=5 when x=-1.
3. Check: 4-(-1)^3=4-(-1)=5

Work out y when x = 5 and when x = -5. Which graph has both of those points in the right places?

1. x cannot be 0 and y is never 0, so the graph has two parts that never touch the axes.
2. Positive x gives positive y, and negative x gives negative y.
3. So one part has x and y both positive and the other has x and y both negative. This is graph C.

Put the y-coordinate into the equation. For part (b), find the number that cubes to give your answer: it can be negative.

Part (a)

1. 2=\frac{10}{q}
2. So 10\div q=2
3. q=10\div2=5

Part (b)

1. -6=r^3+2
2. r^3=-6-2=-8
3. (-2)^3=-2\times-2\times-2=-8, so r=-2.

Test each equation with the values of x in the table. Only one equation fits all but one of the values.

Part (a)

1. Check each value of x in y=x^3+3: x=-2 gives -8+3=-5, x=0 gives 3, x=1 gives 4, x=2 gives 8+3=11. These match.
2. y=x^2+3 gives 7 when x=2, not 11, and y=3-x^3 gives -5 when x=2, not 11.
3. y=3x+3 gives 6 when x=1, not 4.
4. Only y=x^3+3 matches all but one value, so the table is for y=x^3+3.

Part (b)

1. The value for x=-1 is wrong.
2. (-1)^3=-1
3. y=-1+3=2, not 4.

Each small square on both axes is 0.5. For part (b), divide one value of y by the other.

Part (a)

1. Go up from 2 on the x-axis to the curve, then across to the y-axis.
2. Each small square is 0.5, so y=1.5
3. Check: 3\div2=1.5

Part (b)

1. When x=0.5, y=6. When x=2, y=1.5.
2. 6\div1.5=4
3. n=4

Substitute the coordinates of the point to find c. Then use the full equation with y=61

Part (a)

1. Put x=2 and y=5 into the equation: 5=2^3+c
2. 5=8+c
3. c=5-8=-3

Part (b)

1. The equation is y=x^3-3
2. 61=x^3-3, so x^3=64
3. 4^3=4\times4\times4=64, so x=4.

Put x = 0 into each equation for parts (a) and (b). For part (c), work out (-2)^3 carefully.

Part (a)

1. On the y-axis, x=0.
2. y=\frac{1}{x} has no value when x=0, so its graph never crosses the y-axis.
3. The other three all have a value when x=0.

Part (b)

1. Put x=0 into each equation.
2. y=4-0^3=4, y=-4\times0^3=0, y=0^3-4=-4, and y=\frac{-4}{x} has no value at x=0.
3. So the answer is y=x^3-4

Part (c)

1. (-2)^3=-2\times-2\times-2=-8
2. So the point on the graph with x=-2 is (-2, -8), not (-2, 8).