Sketching Graphs and Exponential Graphs

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

1. The curve crosses the y-axis where x=0
2. y=3(0+4)^2-5=3\times16-5
3. y=48-5=43
4. Answer: (0,\ 43)

Put x=0 into the equation to find where the curve crosses the y-axis, then think about whether y gets bigger or smaller as x increases.

1. At x=0: 1.5^0=1, so y=4\times1=4; the graph goes through (0,\ 4)
2. At x=1: y=4\times1.5=6; the graph goes through (1,\ 6)
3. 1.5>1, so y increases as x increases
4. A is y=6^x (the 4 and 1.5 must not be multiplied first), C is y=4\times1.5^{-x} and D is y=1.5\times4^x
5. Answer: B

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

1. From -7 to -1.5 is a distance of 5.5
2. The other crossing is the same distance on the other side of -1.5
3. -1.5+5.5=4
4. Answer: x=4

Set each bracket equal to zero to find where the curve crosses the x-axis, then look at the sign of the x^3 term to decide which way round the curve goes.

1. y=0 when 1-x=0, x+2=0 or x-3=0
2. So the curve crosses the x-axis at x=1, x=-2 and x=3 (graphs A and C)
3. Multiplying the x terms: (-x)\times x\times x=-x^3, so the x^3 term is negative
4. A negative cubic starts high on the left and ends low on the right
5. Check: at x=0, y=1\times2\times(-3)=-6, below the x-axis
6. Answer: C

Substitute x=0 first: b^0=1, so this point tells you a straight away.

1. At (0,\ 4): b^0=1, so a=4
2. At (1,\ 6): 4\times b=6, so b=\dfrac32
3. At x=-1: y=4\times\left(\dfrac32\right)^{-1}=4\times\dfrac23
4. y=\dfrac83

Substitute the point to get b^{-2}=\dfrac{9}{16}, then remember that a negative power means the reciprocal.

1. b^{-2}=\dfrac{9}{16}
2. So \dfrac{1}{b^2}=\dfrac{9}{16}, which gives b^2=\dfrac{16}{9}
3. b is positive, so b=\sqrt{\dfrac{16}{9}}
4. Answer: b=\dfrac43

Divide both sides by 80 first, then write the fraction you get as a power of 2.

1. 80\times2^{-t}=5
2. 2^{-t}=\dfrac{5}{80}=\dfrac{1}{16}
3. \dfrac{1}{16}=2^{-4}, so -t=-4
4. Answer: t=4

A parabola is symmetrical, so its turning point is exactly halfway between the two roots.

1. The turning point is halfway between the roots: x=\dfrac{-2+6}{2}=2
2. The roots -2 and 6 give y=(x+2)(x-6)
3. At x=2: y=(2+2)(2-6)=4\times(-4)=-16
4. Turning point (2,\ -16)

Each point where the curve crosses the x-axis gives a factor of the quadratic, and the x^2 term must come out as 2x^2.

1. x=-3 gives the factor (x+3)
2. x=\dfrac12 gives the factor (2x-1), which also makes the x^2 term 2x^2
3. y=(x+3)(2x-1)=2x^2-x+6x-3
4. Answer: y=2x^2+5x-3

The two points have the same y-coordinate, so they are reflections of each other in the line of symmetry of the curve.

1. The line of symmetry is halfway between x=-3 and x=5: x=\dfrac{-3+5}{2}=1
2. For y=x^2+px+7 the line of symmetry is x=-\dfrac{p}{2}
3. -\dfrac{p}{2}=1
4. Answer: p=-2

Imagine a horizontal line y=k moving up and down the diagram, and count how many times it meets the curve.

1. The solutions of \mathrm{f}(x)=k are where the line y=k meets the curve
2. A line between y=-5 and y=4 meets the curve three times; above 4 or below -5 it meets it once
3. The line y=4 touches the curve at the maximum point and crosses it once more: two solutions
4. The line y=-5 touches the curve at the minimum point and crosses it once more: two solutions
5. Answer: k=4 or k=-5

The line y=k meets the curve y=\mathrm{f}(x) only once when it passes through the turning point.

1. y=\mathrm{f}(x) is a U-shaped curve, so y=k meets it once only at the minimum point
2. 2x^2+12x+23=2(x^2+6x)+23=2(x+3)^2-18+23
3. =2(x+3)^2+5, so the minimum point is (-3,\ 5)
4. Answer: k=5

Rearrange x^2-x-5=0 so that the left-hand side is exactly x^2+2x-3; the right-hand side is the line.

1. Start from x^2-x-5=0
2. Add 3x to both sides: x^2+2x-5=3x
3. Add 2 to both sides: x^2+2x-3=3x+2
4. The left-hand side is the curve, so draw y=3x+2
5. Answer: y=3x+2

Use the y-axis point to find c first, then substitute (2,\ 0) to find b.

1. At (0,\ -8): c=-8
2. At (2,\ 0): 4+2b-8=0, so b=2
3. y=x^2+2x-8=(x-2)(x+4)
4. y=0 when x=2 or x=-4
5. Answer: A(-4,\ 0)

Write the equation as y=a(x+1)(x-5) and use the point on the y-axis to find a.

1. The roots give y=a(x+1)(x-5)
2. At (0,\ 10): a\times1\times(-5)=10, so a=-2
3. The turning point is halfway between the roots: x=\dfrac{-1+5}{2}=2
4. At x=2: y=-2\times3\times(-3)=18
5. Answer: (2,\ 18)

A parabola is symmetrical about the vertical line through its turning point, so P and Q are the same distance from x=-1. Answer part (a) before attempting part (b).

Part (a)

1. P and Q are symmetrical about x=-1, each 8\div2=4 away
2. So the curve crosses the x-axis at x=-1-4=-5 and x=-1+4=3
3. y=(x+5)(x-3)
4. =x^2-3x+5x-15
5. Answer: y=x^2+2x-15

Part (b)

1. At x=-1: y=(-1)^2+2(-1)-15=-16, so T is (-1,\ -16)
2. Base PQ=8 and the height is the distance from T to the x-axis, 16
3. Area =\dfrac12\times8\times16
4. Answer: 64

Substitute both points to get two equations, then divide one by the other so that p cancels. Answer part (a) before attempting part (b).

Part (a)

1. p\times q^{-1}=4 and p\times q^{-3}=\dfrac49
2. Divide the first equation by the second: \dfrac{q^{-1}}{q^{-3}}=4\div\dfrac49
3. q^{2}=9
4. q is positive, so q=3
5. Answer: q=3

Part (b)

1. p\times3^{-1}=4, so \dfrac{p}{3}=4
2. Answer: p=12

Part (c)

1. \mathrm{f}(x)=12\times3^{-x} gets smaller as x increases
2. The largest values are near x=-2: 12\times3^{2}=108, but x=-2 is not in the domain, so use <
3. The smallest value is at x=1: 12\times3^{-1}=4, which is included
4. Answer: 4\leqslant\mathrm{f}(x)<108

Read off where the graph crosses the y-axis first: at x=0, q^0=1. Answer part (a) before attempting part (c).

Part (a)

1. The graph crosses the y-axis at (0,\ 2), and q^0=1, so p=2
2. The graph passes through (1,\ 3), so 2\times q=3
3. Answer: q=1.5

Part (b)

1. Go across from 6 on the y-axis to the curve
2. Go down to the x-axis and read the value
3. Answer: x=2.7 (accept 2.6 to 2.8)

Part (c)

1. 2\times1.5^{k}=\dfrac{81}{8}
2. 1.5^{k}=\dfrac{81}{16}
3. \left(\dfrac32\right)^{k}=\dfrac{3^4}{2^4}=\left(\dfrac32\right)^4
4. Answer: k=4

Rearrange the equation so that one side is exactly x^2+3x-1; the other side is the line to draw. Answer part (a) before attempting part (b).

Part (a)

1. Expand the bracket: x^2+4x=3
2. So x^2+4x-3=0
3. Write the left-hand side as (x^2+3x-1)+x-2
4. So x^2+3x-1+x-2=0, which gives x^2+3x-1=-x+2
5. The line to draw is y=-x+2

Part (b)

1. The line y=-x+2 goes through (0,\ 2) and (2,\ 0)
2. It crosses the curve twice
3. Read the x-value at each crossing
4. x=0.6 and x=-4.6

Differentiate and set \mathrm{f}'(x)=0 to find the stationary points. Answer part (a) before attempting part (b).

Part (a)

1. \mathrm{f}'(x)=3x^2+6x
2. Stationary points where 3x^2+6x=0, so 3x(x+2)=0
3. x=0 or x=-2
4. A positive cubic has its maximum on the left, at x=-2 (or: \mathrm{f}''(-2)=-6<0)
5. \mathrm{f}(-2)=-8+12-2=2
6. Answer: (-2,\ 2)

Part (b)

1. The other stationary point is at x=0
2. \mathrm{f}(0)=-2
3. Answer: (0,\ -2)

Part (c)

1. The solutions are where the line y=k meets the curve
2. The line meets the curve three times when it lies strictly between the minimum and the maximum
3. At k=2 or k=-2 the line touches a turning point, giving only two solutions
4. Answer: -2<k<2