To solve problem (d), which asks where the two graphs cross and how to find these points without looking at the graphs, we need to find the points of intersection of the two equations:

1. $y - 2x = 3$ (Equation 1)
2. $y - 3 = x^2$ (Equation 2)

Step 1: Solve for $y$ in both equations.

From Equation 1: $y = 2x + 3$

From Equation 2: $y = x^2 + 3$

Step 2: Set the equations equal to each other.

Since both expressions represent $y$, we can set them equal to each other to find where the graphs intersect: $2x + 3 = x^2 + 3$

Step 3: Simplify and solve for $x$.

Cancel out the 3 on both sides: $2x = x^2$

Rearrange the equation: $x^2 - 2x = 0$

Factor the equation: $x(x - 2) = 0$

Thus, $x = 0$ or $x = 2$.

Step 4: Find the corresponding $y$ values.

For $x = 0$: $y = 2(0) + 3 = 3$

For $x = 2$: $y = 2(2) + 3 = 4 + 3 = 7$

Step 5: Points of intersection.

The two graphs intersect at the points $(0, 3)$ and $(2, 7)$.

Conclusion:

The points of intersection of the two graphs are $(0, 3)$ and $(2, 7)$. These can be found algebraically by equating the two equations, solving for $x$, and then finding the corresponding $y$-values.

Would you like further details on any of the steps? Here are some related questions to expand your understanding:

1. How do you find the intersection of two linear or nonlinear equations in general?
2. What is the importance of factoring when solving quadratic equations?
3. How can you confirm the solutions to a system of equations graphically?
4. What is the difference between solving algebraically and solving graphically?
5. How can we interpret the intersection points in the context of real-world problems?

Tip: When solving systems of equations, if one equation is quadratic and the other is linear, the number of intersection points will typically be 0, 1, or 2, depending on how the curves intersect.