determine the maximum value of y=3x$^2$ + 5x - 3

1. Analyze the Question

You are asked to find the maximum value of the quadratic function $y = 3x^2 + 5x - 3$.

2. Explain the Concept

A quadratic function is generally written in the form $y = ax^2 + bx + c$.

• If $a > 0$ (positive), the parabola opens upwards, meaning the function has a minimum value at its vertex.
• If $a < 0$ (negative), the parabola opens downwards, meaning the function has a maximum value at its vertex.

In your equation, $a = 3$. Since $3$ is greater than $0$, this parabola opens upwards. Therefore, the function has a minimum value, not a maximum value.

3. Show the Solution

Let's verify the nature of the vertex using the formula for the $x$-coordinate of the vertex: $x = -\frac{b}{2a}$

Given $a = 3$ and $b = 5$: $x = -\frac{5}{2(3)} = -\frac{5}{6}$

To find the corresponding $y$-value: $y = 3(-\frac{5}{6})^2 + 5(-\frac{5}{6}) - 3$ $y = 3(\frac{25}{36}) - \frac{25}{6} - 3$ $y = \frac{25}{12} - \frac{50}{12} - \frac{36}{12}$ $y = -\frac{61}{12} \approx -5.08$

Because the coefficient of $x^2$ is positive, the graph extends infinitely upwards as $x$ increases or decreases. Thus, there is no "maximum" value for this function.

4. Identify the Answer

Since the function opens upwards, it has a minimum value but no maximum value. Looking at the provided options, none of them represent a valid maximum value for this function. Therefore, the correct choice is that there is no correct option provided.

5. Exam Tip

Always check the sign of the coefficient of $x^2$ before calculating. If you see a positive $x^2$ term, the question asks for a "maximum" to test if you know the difference between a vertex that is a minimum and one that is a maximum. Don't fall for the trap of calculating a value when the function doesn't have one!