The task is to determine in how many points a given parabola intersects with a given straight line. The problem presents three pairs of a parabola and a straight line (labeled as p and g, respectively). To solve this, we need to set the equations of the parabola and the line equal to each other and solve for the values of $x$. The number of real solutions tells us the number of intersection points.

Let's solve each case one by one.

a)

Parabola: $p: y = \frac{1}{2}x^2 - 2$
Line: $g: y = x + 4$

Set the two equations equal to find the intersection points: $\frac{1}{2}x^2 - 2 = x + 4$

b)

Parabola: $p: y = -2x^2 + 7$
Line: $g: y = -2x + 3$

Set the two equations equal to find the intersection points: $-2x^2 + 7 = -2x + 3$

c)

Parabola: $p: y = -\frac{1}{4}x^2 + 3$
Line: $g: y = -\frac{1}{2}x + 6$

Set the two equations equal to find the intersection points: $-\frac{1}{4}x^2 + 3 = -\frac{1}{2}x + 6$

I'll now solve each of these equations to find the number of intersection points.The results show the number of intersection points for each case:

• a) The parabola $p: y = \frac{1}{2}x^2 - 2$ and the line $g: y = x + 4$ intersect at 2 points (approximate solutions: $x = -2.61$ and $x = 4.61$).
• b) The parabola $p: y = -2x^2 + 7$ and the line $g: y = -2x + 3$ intersect at 2 points (exact solutions: $x = -1$ and $x = 2$).
• c) The parabola $p: y = -\frac{1}{4}x^2 + 3$ and the line $g: y = -\frac{1}{2}x + 6$ have 0 real intersection points (the solutions are complex numbers, indicating no real intersection).

Summary:

• Case a: 2 intersection points
• Case b: 2 intersection points
• Case c: 0 real intersection points

Would you like further details on these solutions?

Here are some related questions to expand on this topic:

1. What happens when a parabola and a line intersect at exactly one point?
2. How can we visualize the intersection of two functions graphically?
3. What determines whether a quadratic equation has real or complex solutions?
4. Can the line ever lie entirely within or outside the parabola without intersection?
5. How does the discriminant of a quadratic equation relate to the number of intersection points?

Tip: When solving quadratic equations, the discriminant $\Delta = b^2 - 4ac$ is key to determining whether the solutions are real (if $\Delta \geq 0$) or complex (if $\Delta < 0$).