To find the intersection points of two equations, we need to set them equal to each other and solve for $x$. Let's solve the system of equations:

1. $y = mx + c$ (a straight line)
2. $y = ax^2 + bx + c$ (a quadratic equation)

Step-by-Step Solution:

1. Set the two equations equal to each other:
   Since both expressions represent $y$, we can set them equal:

$mx + c = ax^2 + bx + c$

2. Simplify the equation:
   Subtract $c$ from both sides:

$mx = ax^2 + bx$

3. Rearrange the equation:
   Move all terms to one side to get a standard quadratic form:

$0 = ax^2 + (b - m)x$

4. Factor the equation (if possible) or use the quadratic formula:
   The equation is now:

$ax^2 + (b - m)x = 0$

You can factor out $x$:

$x(ax + (b - m)) = 0$

Thus, the solutions are:

$x = 0 \quad \text{or} \quad ax + (b - m) = 0$

5. Solve for $x$:
   For the second case:

$ax = m - b$ $x = \frac{m - b}{a} \quad \text{(if } a \neq 0\text{)}$

Conclusion:

The solutions for $x$ are:

1. $x = 0$
2. $x = \frac{m - b}{a}$ (if $a \neq 0$)

Now, substitute these $x$-values into either original equation (e.g., $y = mx + c$) to find the corresponding $y$-coordinates.

• For $x = 0$: $y = m(0) + c = c$ So one intersection point is $(0, c)$.

• For $x = \frac{m - b}{a}$: $y = m \left(\frac{m - b}{a}\right) + c = \frac{m(m - b)}{a} + c$ So the second intersection point is $\left(\frac{m - b}{a}, \frac{m(m - b)}{a} + c\right)$.

Do you need further details or have any questions?

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Here are 5 related questions to expand on this topic:

1. How can you find the intersection point when dealing with different types of curves (e.g., circles and lines)?
2. How does the quadratic formula apply to more complex quadratic equations?
3. What happens if the discriminant is negative in a quadratic equation?
4. How would you solve if the quadratic and linear equations represent parallel lines?
5. Can you solve for intersections graphically? How would you plot both equations?

Tip: Always check for common factors when solving algebraic equations, as this can simplify the process.