To find the points of intersection between the graphs of the functions $f(x) = (x+5)(x-2)$ and $g(x) = (2x+1)(x-2)$, we need to set the two functions equal to each other and solve for $x$.

Step 1: Set the equations equal to each other.

$(x+5)(x-2) = (2x+1)(x-2)$

Since both sides contain $(x-2)$, we can divide both sides of the equation by $(x-2)$ as long as $x \neq 2$.

$x+5 = 2x+1$

Step 2: Solve for $x$.

$x+5 = 2x+1$ Subtract $x$ from both sides: $5 = x+1$ Subtract 1 from both sides: $x = 4$

So, one point of intersection occurs when $x = 4$.

Step 3: Find the corresponding $y$-coordinate.

Substitute $x = 4$ into either function to find the $y$-value. Let's use $f(x)$: $f(4) = (4+5)(4-2) = 9 \times 2 = 18$ Thus, the point of intersection is $(4, 18)$.

Step 4: Check if $x = 2$ is also a solution.

Since both functions have $(x-2)$ as a factor, let's substitute $x = 2$ into both functions: $f(2) = (2+5)(2-2) = 7 \times 0 = 0$ $g(2) = (2 \times 2 + 1)(2-2) = 5 \times 0 = 0$ Both functions give $y = 0$ when $x = 2$. Therefore, $(2, 0)$ is also a point of intersection.

Step 5: Conclusion

The points of intersection are $(4, 18)$ and $(2, 0)$.

The correct selections are:

• D) $(2, 0)$
• E) $(4, 18)$

Would you like more details on the calculations or further explanation?

Here are 5 related questions to explore:

1. How do you factor quadratic equations like $f(x) = (x+5)(x-2)$?
2. How would the points of intersection change if the functions were modified?
3. What is the significance of dividing by $(x-2)$ in this problem?
4. Can there be cases where dividing by a common factor might lose valid solutions?
5. How can you graphically verify the points of intersection?

Tip: Always check if common factors can affect the domain of your solution before dividing through them!