The task requires determining the intersection points of the given functions $f(x)$ with the coordinate axes, i.e., finding where the functions intersect with the x-axis (roots) and the y-axis (when $x = 0$).

Let’s go through each function and find these intersections:

1. Intersection with the y-axis: Set $x = 0$ to find $f(0)$.
2. Intersection with the x-axis: Set $f(x) = 0$ and solve for $x$ (this gives the roots).

a) $f(x) = -1.5x^2 - 6$

• Y-axis: $f(0) = -1.5(0)^2 - 6 = -6$ → Intersection at $(0, -6)$.
• X-axis: Solve $-1.5x^2 - 6 = 0$, $-1.5x^2 = 6$, $x^2 = -4$ → No real roots (no intersection with x-axis).

b) $f(x) = x^2 + 5x$

• Y-axis: $f(0) = (0)^2 + 5(0) = 0$ → Intersection at $(0, 0)$.
• X-axis: $x^2 + 5x = 0$, $x(x + 5) = 0$ → Roots $x = 0$ and $x = -5$ → Intersections at $(0, 0)$ and $(-5, 0)$.

c) $f(x) = x^2 - 4x - 45$

• Y-axis: $f(0) = (0)^2 - 4(0) - 45 = -45$ → Intersection at $(0, -45)$.
• X-axis: Solve $x^2 - 4x - 45 = 0$, factor as $(x - 9)(x + 5) = 0$, roots $x = 9$ and $x = -5$ → Intersections at $(9, 0)$ and $(-5, 0)$.

d) $f(x) = (x + 1.5)(x + 3.5)$

• Y-axis: $f(0) = (0 + 1.5)(0 + 3.5) = 1.5 \times 3.5 = 5.25$ → Intersection at $(0, 5.25)$.
• X-axis: Solve $(x + 1.5)(x + 3.5) = 0$, roots $x = -1.5$ and $x = -3.5$ → Intersections at $(-1.5, 0)$ and $(-3.5, 0)$.

e) $f(x) = x^2 - 3x + 2$

• Y-axis: $f(0) = (0)^2 - 3(0) + 2 = 2$ → Intersection at $(0, 2)$.
• X-axis: Solve $x^2 - 3x + 2 = 0$, factor as $(x - 1)(x - 2) = 0$, roots $x = 1$ and $x = 2$ → Intersections at $(1, 0)$ and $(2, 0)$.

f) $f(x) = \frac{1}{2}x^2 + 4x - 10$

• Y-axis: $f(0) = \frac{1}{2}(0)^2 + 4(0) - 10 = -10$ → Intersection at $(0, -10)$.
• X-axis: Solve $\frac{1}{2}x^2 + 4x - 10 = 0$, multiply through by 2: $x^2 + 8x - 20 = 0$. Use the quadratic formula: $x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-20)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 80}}{2} = \frac{-8 \pm \sqrt{144}}{2} = \frac{-8 \pm 12}{2}$ Roots: $x = 2$ and $x = -10$ → Intersections at $(2, 0)$ and $(-10, 0)$.

g) $f(x) = (x - 2)^2 - 1$

• Y-axis: $f(0) = (0 - 2)^2 - 1 = 4 - 1 = 3$ → Intersection at $(0, 3)$.
• X-axis: Solve $(x - 2)^2 - 1 = 0$, $(x - 2)^2 = 1$, $x - 2 = \pm 1$, so $x = 3$ and $x = 1$ → Intersections at $(3, 0)$ and $(1, 0)$.

Let me know if you want further clarifications or have any specific questions.

Here are 5 relative questions to expand your understanding:

1. How can we determine the number of intersections a quadratic function has with the x-axis?
2. What is the significance of the discriminant in finding the roots of quadratic equations?
3. How does factoring help in finding the roots of polynomial functions?
4. How do transformations of basic quadratic functions affect their intersections with the axes?
5. Can a quadratic function have no real intersections with the x-axis? Why?

Tip: For quadratic equations, always check if factoring is possible before using the quadratic formula—it often simplifies the solution!