Quadratic Graphs

Put x=2 into the equation. Work out x^2 and 2x separately.

Squaring a negative number gives a positive answer, and subtracting a negative is the same as adding.

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

A quadratic graph is a smooth curve with one turning point. Think about which way up it is when the x^2 term is positive.

The turning point is where the curve stops going down and starts going up.

The line of symmetry is a vertical line through the lowest point of the curve. A vertical line has an equation x = \ldots

1. The two sides of the curve reflect in a vertical line through the turning point (-1,-4).

2. Every point on this vertical line has x=-1, so its equation is x=-1.

Look for the points where the curve meets the horizontal axis. There are two of them.

1. The graph crosses the x-axis where y=0.
2. It crosses at (-1, 0) and (4, 0).
3. So x=-1 and x=4.

Points the same distance either side of the turning point have the same value of y.

1. The turning point is at x=2, so the graph is symmetrical about the line x=2.
2. x=0 and x=4 are each 2 units from x=2, on opposite sides.

3. The point at x=0 has y=5. Its matching point at x=4 also has y=5, so the missing value is 5.

Check one of Sam's points. If the points are right, think about how they are joined.

1. Sam's points are correct. For example, when x=1, y=1^2-4\times1+1=-2.
2. The graph of y=x^2-4x+1 is a smooth U-shaped curve, not a set of straight lines.
3. The points should be joined with a smooth curve.

Put the value of x into the equation. Work out x^2 first, then 2x.

Part (a)

1. y=3^2-2\times3-1
2. y=9-6-1=2
3. The point is (3, 2).

Part (b)

1. y=1^2-2\times1-1
2. y=1-2-1=-2
3. The point is (1, -2).

Work out each term separately. Squaring a negative number gives a positive answer.

Part (a)

1. (-1)^2=1 and 5\times(-1)=-5
2. y=1-(-5)+3=1+5+3=9

Part (b)

1. y=0^2-5\times0+3=0-0+3=3

Part (c)

1. y=2^2-5\times2+3
2. y=4-10+3=-3

When the x^2 term is negative, the turning point is the highest point of the curve.

1. This graph is an upside-down U shape, so the turning point is the highest point.
2. The highest point is at x=1 and y=7.
3. The coordinates are (1, 7).

Is the x^2 term positive or negative? Then put x=0 into the equation.

1. The x^2 term is negative, so the graph is an upside-down U shape. That is graph B or graph C.
2. When x=0, y=4-0=4, so the graph crosses the y-axis above O.
3. This is graph C.

The two points are the same distance from the line of symmetry, one on each side.

1. The two crossings have the same height, y=0, so they are reflections in the line x=2.

2. The given point is 2-(-1)=3 units to the left of x=2. The other crossing is 3 units to the right: 2+3=5.
3. It is still on the x-axis, so its y-coordinate is 0. The other point is (5,0).

Put y = 11 into the equation. There are two numbers that square to give the same answer.

1. At P, y=11, so k^2+2=11
2. k^2=9, so k=3 or k=-3.
3. P is to the left of the y-axis, so its x-coordinate is negative.
4. k=-3

Find two values of x with the same value of y. The turning point is halfway between them.

Part (a)

1. The table gives y=-1 at both x=-2 and x=-1. These two points are at the same height.

2. The turning point lies on the line of symmetry, halfway between these x-values: \frac{-2+(-1)}{2}=-1.5.
3. The x-coordinate of the turning point is -1.5.

Part (b)

1. Put x=-1.5 into the equation.
2. (-1.5)^2=2.25 and 3\times(-1.5)=-4.5
3. y=2.25-4.5+1=-1.25

The graph is symmetrical about the line through its lowest point. Look for a point that does not match the point on the other side.

Part (a)

1. (-1)^2=1 and 2\times(-1)=-2
2. y=1-(-2)-5=1+2-5=-2

Part (b)

1. Check each point. When x=3, y=3^2-2\times3-5=9-6-5=-2
2. Ali plotted (3, 2), which is wrong. The other five points are correct.
3. The point should be at (3, -2).
4. This also fits the symmetry: x=-1 and x=3 are the same distance from x=1, so they have the same value of y.

For part (b), find the two times when the height is exactly 8 metres. Each small square across is 0.5 seconds.

Part (a)

1. The greatest height is at the top of the curve.
2. The top of the curve is at h=9, so the greatest height is 9 metres.

Part (b)

1. Find where the curve is at a height of 8 metres.
2. The curve is at h=8 when t=1 and when t=3.
3. Between these times the ball is higher than 8 metres.
4. 3-1=2 seconds

Points with the same value of y are the same distance either side of the line of symmetry.

Part (a)

1. The points have the same y-coordinate, 6, so the line of symmetry is halfway between them.

2. Halfway between x=-1 and x=5 is \frac{-1+5}{2}=2.
3. The line of symmetry is x=2.

Part (b)

1. The point (7,22) is 7-2=5 units to the right of x=2. Reflect it 5 units to the left to keep the same height.

2. The other x-coordinate is 2-5=-3.

Substitute the coordinates of the point into the equation, then solve for b.

Part (a)

1. Put x=2 and y=-5 into the equation: -5=2^2+2b-1
2. -5=4+2b-1, so -5=3+2b
3. 2b=-8, so b=-4.

Part (b)

1. The equation is y=x^2-4x-1
2. y=5^2-4\times5-1=25-20-1=4

Part (c)

1. y=1^2-4\times1-1=1-4-1=-4