Answer:
The equation of the parabola is [tex]y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}[/tex]. The average rate of change of the parabola is -4.
Step-by-step explanation:
We must remember that a parabola is represented by a quadratic function, which can be formed by knowing three different points. A quadratic function is standard form is represented by:
[tex]y = a\cdot x^{2}+b\cdot x + c[/tex]
Where:
[tex]x[/tex] - Independent variable, dimensionless.
[tex]y[/tex] - Dependent variable, dimensionless.
[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Coefficients, dimensionless.
If we know that [tex](3, -30)[/tex], [tex](-2, 0)[/tex] and [tex](18, 0)[/tex] are part of the parabola, the following linear system of equations is formed:
[tex]9\cdot a +3\cdot b + c = -30[/tex]
[tex]4\cdot a -2\cdot b +c = 0[/tex]
[tex]324\cdot a +18\cdot b + c = 0[/tex]
This system can be solved both by algebraic means (substitution, elimination, equalization, determinant) and by numerical methods. The solution of the linear system is:
[tex]a = \frac{2}{5}[/tex], [tex]b = -\frac{32}{5}[/tex], [tex]c = -\frac{72}{5}[/tex].
The equation of the parabola is [tex]y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}[/tex].
Now, we calculate the average rate of change ([tex]r[/tex]), dimensionless, between [tex]x = -2[/tex] and [tex]x = 8[/tex] by using the formula of secant line slope:
[tex]r = \frac{y(8)-y(-2)}{8-(-2)}[/tex]
[tex]r = \frac{y(8)-y(-2)}{10}[/tex]
[tex]x = -2[/tex]
[tex]y = \frac{2}{5}\cdot (-2)^{2}-\frac{32}{5}\cdot (-2)-\frac{72}{5}[/tex]
[tex]y(-2) = 0[/tex]
[tex]x = 8[/tex]
[tex]y = \frac{2}{5}\cdot (8)^{2}-\frac{32}{5}\cdot (8)-\frac{72}{5}[/tex]
[tex]y(8) = -40[/tex]
[tex]r = \frac{-40-0}{10}[/tex]
[tex]r = -4[/tex]
The average rate of change of the parabola is -4.
