Recognising and Sketching Graphs

Is the curve U-shaped, S-shaped or in two separate pieces?

1. The curve is U-shaped, symmetrical about the y-axis, with its lowest point at the origin. This is y=x^2.
2. y=-x^2 is the same shape upside down, y=x^3 is below the x-axis for negative x, and y=\frac1x never meets the axes.

Is y negative or positive when x is negative?

1. The curve passes through the origin. When x is negative y is negative, and when x is positive y is positive, so it rises from bottom left to top right.
2. This is the cubic y=x^3; for example, x=-1 gives y=-1.
3. y=-x^3 falls from top left to bottom right, y=x^2 is U-shaped, and y=2^x is never below the x-axis.

A graph in two separate pieces comes from dividing by x.

1. The graph is in two separate pieces that get closer and closer to the axes without touching them: a reciprocal graph, y=\frac kx.
2. When x is positive y is positive (top right), and when x is negative y is negative (bottom left), so k is positive: y=\frac2x.
3. y=-\frac2x would have its pieces in the top left and bottom right.

Is y ever positive on this graph?

1. The curve is an upside-down U, symmetrical about the y-axis, with its highest point at the origin.
2. Every y-value is 0 or negative. That fits y=-x^2: x^2 is never negative, so -x^2 is never positive.
3. y=x^2 is the same shape the other way up.

Try x=-1 in each equation and look at the sign of y.

1. The curve passes through the origin and falls from top left to bottom right: y is positive for negative x and negative for positive x.
2. That is y=-x^3; for example, x=-1 gives y=-(-1)^3=1.
3. y=-x is a straight line, y=-x^2 is an upside-down U, and y=x^3 rises from bottom left to top right.

Look at the sign of y when x is positive.

1. The graph is in two separate pieces approaching the axes, so it is a reciprocal graph.
2. For negative x the curve is above the x-axis, and for positive x it is below, so x and y always have opposite signs.
3. That needs a negative number on top: y=-\frac2x (for example, x=1 gives y=-2). y=\frac2x has its pieces in the top right and bottom left.

Where does the curve cross the y-axis?

1. The curve is always above the x-axis, crosses the y-axis at 1 and rises more and more steeply: an exponential graph.
2. y=2^x fits: 2^0=1, and y doubles each time x goes up by 1.
3. y=x^2 and y=x^3 pass through the origin, and y=2x is a straight line.

Work out the value of each equation when x=0.

1. The curve is always above the x-axis, crosses the y-axis at 1 and falls as x increases: a decreasing exponential graph.
2. y=2^{-x} is the same as y=\left(\frac12\right)^x: it gives 1 when x=0 and halves each time x goes up by 1.
3. y=2^x rises instead, y=-2^x is below the x-axis, and y=x^{-1}=\frac1x is in two separate pieces.

Find where the graph crosses the y-axis, then decide whether it is U-shaped or upside down.

1. When x=0, y=4-0=4, so the graph crosses the y-axis at 4.
2. The -x^2 makes the curve an upside-down U, so (0,4) is its highest point.
3. Graph C has both features (it crosses the x-axis where x^2=4, at -2 and 2).
4. Graph A is y=x^2-4: the right x-axis crossings, but U-shaped with its lowest point at (0,-4).

The turning point is halfway between the two points where the curve crosses the x-axis.

1. The curve crosses the x-axis where y=0: x-3=0 or x+1=0, so at x=3 and x=-1.
2. A quadratic curve is symmetrical, so the turning point is halfway between: x=\frac{3+(-1)}{2}=1.
3. Substitute x=1: y=(1-3)(1+1)=-2\times2=-4.
4. The turning point is (1,-4), the lowest point of the curve.

y=0 on the x-axis: read the x-coordinates where the curve crosses it.

1. y=0 at every point on the x-axis, so look where the curve crosses the x-axis.
2. It crosses at about x=-2 and x=3.

Imagine the horizontal line y=3 and read the x-coordinates where it meets the curve.

1. Draw the horizontal line y=3 across the grid.
2. It meets the curve twice, at points the same distance either side of the highest point, x=2.
3. Reading down from these points gives x\approx0.6 and x\approx3.4.

Read the lowest point of the curve, giving the x-coordinate first.

1. The minimum is at approximately x=1 and y=-2.
2. The turning point is (1,-2).

y=0 on the x-axis: read the x-coordinates where the curve crosses it.

1. y=0 on the x-axis, so read where the curve crosses it.
2. The three crossings are at about x=-3, x=1 and x=3.

Substitute the coordinates; a negative power means a reciprocal.

1. Substitute x=-2 and y=6.25: k^{-2}=6.25.
2. A negative power is a reciprocal, so \frac1{k^2}=6.25=\frac{25}4.
3. Turn both sides upside down: k^2=\frac4{25}.
4. k is positive, so k=\sqrt{\frac4{25}}=\frac25=0.4.

Use where the curve crosses the x-axis and where it crosses the y-axis.

1. The curve crosses the x-axis at -2 and 1, so its equation is y=a(x+2)(x-1) for some number a.
2. It crosses the y-axis at -4. When x=0, a(0+2)(0-1)=-2a, so -2a=-4.
3. So a=2 and the equation is y=2(x+2)(x-1).
4. y=(x+2)(x-1) has the right roots but crosses the y-axis at -2; y=-2(x+2)(x-1) would be an upside-down U.

Complete the square on x^2-4x, then write y as a number minus a square.

Part (a)

1. Complete the square: x^2-4x=(x-2)^2-4.
2. So 7+4x-x^2=7-(x^2-4x)=7-\big((x-2)^2-4\big)=11-(x-2)^2.
3. (x-2)^2 is never negative, so y is largest when (x-2)^2=0: at x=2, where y=11.
4. The turning point is (2,11), a maximum.

Part (b)

1. On the x-axis y=0, so 11-(x-2)^2=0, which gives (x-2)^2=11.
2. Take the square root of both sides, remembering both signs: x-2=\pm\sqrt{11}.
3. So x=2+\sqrt{11} or x=2-\sqrt{11}.

Substitute each point to get two equations in p and q.

1. Substitute (-1,2): 1-p+q=2, so q-p=1.
2. Substitute (3,-6): 9+3p+q=-6, so 3p+q=-15.
3. Subtract the first equation from the second: 4p=-16, so p=-4. Then q=1+p=-3.
4. So y=x^2-4x-3. Complete the square: x^2-4x-3=(x-2)^2-4-3=(x-2)^2-7.
5. The turning point is (2,-7).

Find k first, then solve the equations of the curve and the line together.

1. P(2,3) is on the curve, so 3=\frac k2 and k=6. The curve is y=\frac6x.
2. Where the line meets the curve, x+1=\frac6x. Multiply both sides by x: x^2+x=6.
3. Rearrange and factorise: x^2+x-6=0, so (x+3)(x-2)=0.
4. x=2 is the point P, so at Q, x=-3 and y=-3+1=-2.
5. Q is (-3,-2).

Write an equation for each point, then divide one by the other so that a cancels.

Part (a)

1. Substitute each point into y=ab^x: ab=10 and ab^3=40.
2. Divide the second equation by the first. The a cancels: \frac{ab^3}{ab}=\frac{40}{10}, so b^2=4.
3. b is positive, so b=2.

Part (b)

1. Use ab=10 with b=2: 2a=10, so a=5.
2. Check: y=5\times2^x gives 5\times2^3=40 at x=3.