Question 1 – Finding Intersection with Coordinate Axes

In this question, we have four linear functions. For each, we need to find where the graph meets the x-axis and y-axis.

The y-axis is where x = 0. So to find the y-intercept, we put x = 0 in the equation and solve for y. This gives us the point (0, y).

The x-axis is where y = 0. So to find the x-intercept, we put y = 0 in the equation and solve for x. This gives us the point (x, 0).

We plot these two points and draw a straight line through them. The line will cross the axes at these points.

For part (i) y = –5x + 10:

• y-intercept: put x = 0 → y = 10 → point (0, 10)

• x-intercept: put y = 0 → 0 = –5x + 10 → 5x = 10 → x = 2 → point (2, 0)

For part (ii) y = 2x – 1:

• y-intercept: (0, –1)

• x-intercept: 0 = 2x – 1 → 2x = 1 → x = 1/2 → point (1/2, 0)

For part (iii) y = 12x – 3:

• y-intercept: (0, –3)

• x-intercept: 0 = 12x – 3 → 12x = 3 → x = 1/4 → point (1/4, 0)

For part (iv) y = 3x + 2:

• y-intercept: (0, 2)

• x-intercept: 0 = 3x + 2 → 3x = –2 → x = –2/3 → point (–2/3, 0)

Question 2 – Finding Intersection of Two Functions

In this question, we have eight pairs of functions. For each pair, we need to find the point(s) where their graphs intersect.

The intersection point is where both functions give the same value for the same x. So we set f(x) = g(x) and solve for x. Then we find the y-coordinate by putting x back into either function.

For parts (i) to (iv), both functions are linear. We solve the equation f(x) = g(x) and get one unique solution. This means the two lines intersect at exactly one point.

For parts (v) to (viii), one function is linear and the other is quadratic. When we set them equal, we get a quadratic equation. This may have two solutions, one solution, or no solution. If it has two solutions, the graphs intersect at two points.

Part (v) f(x) = x – 1 and g(x) = x² – 4x + 3:
We set x – 1 = x² – 4x + 3. This gives x² – 5x + 4 = 0. Factor: (x – 1)(x – 4) = 0. So x = 1 and x = 4. Then find y for each: for x = 1, y = 0; for x = 4, y = 3. So points are (1, 0) and (4, 3).

Part (vi) f(x) = 3x + 4 and g(x) = x² + 2x – 8:
Set 3x + 4 = x² + 2x – 8. This gives x² – x – 12 = 0. Factor: (x – 4)(x + 3) = 0. So x = 4 and x = –3. Then y values: for x = 4, y = 16; for x = –3, y = –5. Points are (4, 16) and (–3, –5).

Part (vii) f(x) = –2x – 1 and g(x) = x² – 4x:
Set –2x – 1 = x² – 4x. This gives x² – 2x + 1 = 0. Factor: (x – 1)² = 0. So x = 1 only. Then y = –3. So only one point (1, –3), which means the line is tangent to the parabola.

Part (viii) f(x) = –x² – 3x + 2 and g(x) = x + 6:
Set –x² – 3x + 2 = x + 6. This gives –x² – 4x – 4 = 0, or x² + 4x + 4 = 0. Factor: (x + 2)² = 0. So x = –2 only. Then y = 4. So one point (–2, 4), meaning tangent.

Question 3 – Graphing Functions

In this question, we have several functions to graph. The functions involve square roots and quadratics.

For square root functions, we need to find the domain first. The expression under the square root must be greater than or equal to zero. Then we plot points by choosing x values in the domain and finding corresponding y values.

Part (i) y = √(3x): Domain is x ≥ 0. Plot points like (0, 0), (1, √3), (4, √12).

Part (ii) y = √(x + 5): Domain is x + 5 ≥ 0, so x ≥ –5. Plot points like (–5, 0), (–4, 1), (–1, 2).

Part (iii) y = –√x: Domain is x ≥ 0. This is the reflection of √x across the x-axis. Points like (0, 0), (1, –1), (4, –2).

Part (iv) y = –√(x + 1): Domain is x ≥ –1. This is the reflection of √(x + 1) across the x-axis.

Part (v) y = √(2x + 1): Domain is 2x + 1 ≥ 0, so x ≥ –1/2.

Part (vi) y = x² + x – 2: This is a quadratic. It opens upward because coefficient of x² is positive. Find vertex, y-intercept, x-intercepts.

Part (vii) y = x² + x – 2 is repeated? Actually the textbook has (vi) and (vii) with different functions, but the given list shows (vi) as something else. I will explain the general method for quadratic graphs.

For quadratic functions like y = x² + x – 2, we find:

• y-intercept: put x = 0 → y = –2 → point (0, –2)

• x-intercepts: solve x² + x – 2 = 0 → (x + 2)(x – 1) = 0 → x = –2 and x = 1 → points (–2, 0) and (1, 0)

• Vertex: x-coordinate is –b/(2a) = –1/2. Then y = (–1/2)² + (–1/2) – 2 = 1/4 – 1/2 – 2 = –9/4. So vertex is (–1/2, –9/4).

Then we plot these points and draw a smooth parabola.

Question 4 – Building and Tree Height Problem

We have two functions:

• Building height: H(t) = 100 + 20t (metres) where t is in months.

• Tree height: T(t) = 50 + 10t + t²? Actually the question shows T(t) = 50 + 10t + 2? Wait, it says “T()=50+10+2” which seems incomplete. I think it should be T(t) = 50 + 10t + t² or something similar. The textbook shows T(t) = 50 + 10t + 2? Actually it says “T()=50+10+2” – I think there might be a typo. Probably it is T(t) = 50 + 10t + t²/2 or something. Let me explain the general method.

Part (i): To find when both have the same height, we set H(t) = T(t) and solve for t.

Part (ii): That height will be the value of H(t) or T(t) at that time.

We also need to sketch both graphs and determine when the tree overtakes the building. This means the time when T(t) becomes greater than H(t).

Graphically, we draw both curves. The tree starts lower but grows faster (quadratic) while the building grows linearly. So initially, building is higher. Then they meet at some point. After that, the tree becomes higher. We find this time.

Question 5 – Radioactive Decay Problem

The question says: “A radioactive substance has a half-life of 2 years. If the initial quantity is 200 grams and the exponential decay function is Q(t) = … then find the remaining quantity after 6 years graphically.”

The exponential decay function is Q(t) = Q₀(1/2)^(t/T), where T is the half-life. Here Q₀ = 200 grams and T = 2 years.

So Q(t) = 200(1/2)^(t/2).

We need to find the remaining quantity after 6 years. Put t = 6: Q(6) = 200(1/2)^(6/2) = 200(1/2)^3 = 200 × (1/8) = 25 grams.

We can also find this graphically by plotting the decay curve and reading the value at t = 6.

To plot, we find values at t = 0, 2, 4, 6, 8:

• t = 0: Q = 200

• t = 2: Q = 100

• t = 4: Q = 50

• t = 6: Q = 25

• t = 8: Q = 12.5

We plot these points and draw a smooth curve. The curve will show exponential decay.