Intersection of y = x² - 3x and y = 2x - 0.5x² at x=0, 3 1/3

Answer

The intersection points of two functions are found by setting their equations equal to each other. By equating $x^{2} - 3x$ and $2x - \frac{1}{2}x^{2}$ and solving the resulting quadratic equation, we confirm that the graphs meet at $x = 0$ and $x = \frac{10}{3}$ (which is $3 \frac{1}{3}$).

Explanation

Observation of the Image: The image displays a Cartesian plane with two parabolas. One parabola opens upwards ($y = x^2 - 3x$), representing a positive leading coefficient, and the other opens downwards ($y = 2x - \frac{1}{2}x^2$), representing a negative leading coefficient. They intersect at the origin $(0,0)$ and at a point in the first quadrant where $x$ is slightly greater than $3$.

1. Equate the functions To find where two graphs meet, we set the $y$-values equal to each other because, at the point of intersection, both functions share the same coordinates. $x^{2} - 3x = 2x - \frac{1}{2}x^{2}$ This equation represents the condition where the height of the first curve is identical to the height of the second curve.

2. Rearrange into standard quadratic form Move all terms to one side of the equation to set it to zero. Add $\frac{1}{2}x^{2}$ to both sides and subtract $2x$ from both sides. $x^{2} + \frac{1}{2}x^{2} - 3x - 2x = 0$ Grouping like terms allows us to simplify the relationship into a single quadratic expression. $\frac{3}{2}x^{2} - 5x = 0$ This formula shows the combined quadratic equation that must be solved to find the roots (intersections). ⚠️ This step is required on exams to demonstrate algebraic manipulation.

3. Solve by factoring Since there is no constant term, we can factor out the greatest common factor, which is $x$. $x \left( \frac{3}{2}x - 5 \right) = 0$ Factoring converts a polynomial expression into a product of linear factors to identify where the expression equals zero.

4. Identify the roots According to the Zero Product Property, if the product of two factors is zero, at least one of the factors must be zero. Set the first factor to zero: $x_{1} = 0$ Set the second factor to zero: $\frac{3}{2}x - 5 = 0$ This gives us the logical pathways to determine separate intersection points.

5. Solve for the second x-value Isolate $x$ in the second linear equation. $\frac{3}{2}x = 5$ $x = 5 \cdot \frac{2}{3}$ $x_{2} = \frac{10}{3} = 3 \frac{1}{3}$ Multiplying by the reciprocal isolates the variable to find the exact horizontal position of the second intersection.

Final Answer

The intersection points occur at: $\boxed{x = 0 \text{ and } x = 3 \frac{1}{3}}$

Common Mistakes

• Dividing by $x$: Students often divide both sides of $\frac{3}{2}x^2 - 5x = 0$ by $x$. This is a major error because it results in the loss of the $x=0$ solution. Always factor instead.
• Sign Errors: When moving terms between sides (e.g., moving $-\frac{1}{2}x^2$ to the left), forgetting to change the sign will lead to an incorrect quadratic equation.