To solve the problem of finding the changes of sign and extreme values for the given quadratic expressions, we will follow these steps:
### i) For the expression \(15 + 4x - 3x^2\)
1. **Identify coefficients**:
We rewrite the expression in standard form \(ax^2 + bx + c\):
\[
-3x^2 + 4x + 15
\]
Here, \(a = -3\), \(b = 4\), and \(c = 15\).
2. **Calculate the discriminant**:
The discriminant \(\Delta\) is given by \(b^2 - 4ac\):
\[
\Delta = 4^2 - 4 \cdot (-3) \cdot 15 = 16 + 180 = 196
\]
Since \(\Delta > 0\), the quadratic has two distinct real roots.
3. **Finding the roots**:
We use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-4 \pm \sqrt{196}}{2 \cdot -3} = \frac{-4 \pm 14}{-6}
\]
This gives us two roots:
\[
x_1 = \frac{10}{-6} = -\frac{5}{3}, \quad x_2 = \frac{-18}{-6} = 3
\]
4. **Determine the intervals**:
The roots divide the number line into intervals:
- \( (-\infty, -\frac{5}{3}) \)
- \( (-\frac{5}{3}, 3) \)
- \( (3, \infty) \)
5. **Test the sign in each interval**:
- For \(x < -\frac{5}{3}\), choose \(x = -2\):
\[
15 + 4(-2) - 3(-2)^2 = 15 - 8 - 12 = -5 \quad (\text{negative})
\]
- For \(-\frac{5}{3} < x < 3\), choose \(x = 0\):
\[
15 + 4(0) - 3(0)^2 = 15 \quad (\text{positive})
\]
- For \(x > 3\), choose \(x = 4\):
\[
15 + 4(4) - 3(4)^2 = 15 + 16 - 48 = -17 \quad (\text{negative})
\]
6. **Conclusion on sign changes**:
The expression changes sign at \(x = -\frac{5}{3}\) and \(x = 3\). It is negative in the intervals \((- \infty, -\frac{5}{3})\) and \((3, \infty)\), and positive in \((- \frac{5}{3}, 3)\).
7. **Finding the extreme value**:
Since \(a < 0\), the quadratic opens downwards, indicating a maximum at the vertex:
\[
x = -\frac{b}{2a} = -\frac{4}{2 \cdot -3} = \frac{2}{3}
\]
Substitute \(x = \frac{2}{3}\) back into the expression to find the maximum value:
\[
15 + 4\left(\frac{2}{3}\right) - 3\left(\frac{2}{3}\right)^2 = 15 + \frac{8}{3} - 3\left(\frac{4}{9}\right) = 15 + \frac{8}{3} - \frac{4}{3} = 15 + \frac{4}{3} = \frac{45}{3} + \frac{4}{3} = \frac{49}{3}
\]
### ii) For the expression \(4x - 5x^2 + 2\)
1. **Identify coefficients**:
Rewrite in standard form:
\[
-5x^2 + 4x + 2
\]
Here, \(a = -5\), \(b = 4\), and \(c = 2\).
2. **Calculate the discriminant**:
\[
\Delta = 4^2 - 4 \cdot (-5) \cdot 2 = 16 + 40 = 56
\]
Since \(\Delta > 0\), there are two distinct real roots.
3. **Finding the roots**:
Using the quadratic formula:
\[
x = \frac{-4 \pm \sqrt{56}}{2 \cdot -5} = \frac{-4 \pm 2\sqrt{14}}{-10} = \frac{2 \pm \sqrt{14}}{5}
\]
Let \(x_1 = \frac{2 - \sqrt{14}}{5}\) and \(x_2 = \frac{2 + \sqrt{14}}{5}\).
4. **Determine the intervals**:
The roots divide the number line into intervals:
- \( (-\infty, x_1) \)
- \( (x_1, x_2) \)
- \( (x_2, \infty) \)
5. **Test the sign in each interval**:
- For \(x < x_1\), choose \(x = 0\):
\[
2 \quad (\text{positive})
\]
- For \(x_1 < x < x_2\), choose \(x = 1\):
\[
4 - 5 + 2 = 1 \quad (\text{positive})
\]
- For \(x > x_2\), choose \(x = 2\):
\[
8 - 20 + 2 = -10 \quad (\text{negative})
\]
6. **Conclusion on sign changes**:
The expression changes sign at \(x = x_1\) and \(x = x_2\). It is positive in the intervals \((- \infty, x_1)\) and \((x_1, x_2)\), and negative in \((x_2, \infty)\).
7. **Finding the extreme value**:
Since \(a < 0\), the quadratic opens downwards, indicating a maximum at the vertex:
\[
x = -\frac{b}{2a} = -\frac{4}{2 \cdot -5} = \frac{2}{5}
\]
Substitute \(x = \frac{2}{5}\) back into the expression to find the maximum value:
\[
4\left(\frac{2}{5}\right) - 5\left(\frac{2}{5}\right)^2 + 2 = \frac{8}{5} - 5\left(\frac{4}{25}\right) + 2 = \frac{8}{5} - \frac{20}{25} + 2 = \frac{8}{5} - \frac{4}{5} + \frac{10}{5} = \frac{14}{5}
\]
### Summary of Results:
- For \(15 + 4x - 3x^2\):
- Changes of sign: at \(x = -\frac{5}{3}\) and \(x = 3\).
- Maximum value: \(\frac{49}{3}\).
- For \(4x - 5x^2 + 2\):
- Changes of sign: at \(x = \frac{2 - \sqrt{14}}{5}\) and \(x = \frac{2 + \sqrt{14}}{5}\).
- Maximum value: \(\frac{14}{5}\).
