Question

Write an equation of a parabola that passes through (3,-30) and has x-intercepts of -2 and 18. Then find the average rate of change from x= -2 to x=8.

Answer 1

Answer:

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$.  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:

$y = a\cdot x^{2}+b\cdot x + c$

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$a$, $b$, $c$ - Coefficients, dimensionless.

If we know that $(3, -30)$, $(-2, 0)$ and $(18, 0)$ are part of the parabola, the following linear system of equations is formed:

$9\cdot a +3\cdot b + c = -30$

$4\cdot a -2\cdot b +c = 0$

$324\cdot a +18\cdot b + c = 0$

This system can be solved both by algebraic means (substitution, elimination, equalization, determinant) and by numerical methods. The solution of the linear system is:

$a = \frac{2}{5}$, $b = -\frac{32}{5}$, $c = -\frac{72}{5}$.

The equation of the parabola is $y = \frac{2}{5}\cdot x^{2}-\frac{32}{5}\cdot x -\frac{72}{5}$.

Now, we calculate the average rate of change ($r$), dimensionless, between $x = -2$ and $x = 8$ by using the formula of secant line slope:

$r = \frac{y(8)-y(-2)}{8-(-2)}$

$r = \frac{y(8)-y(-2)}{10}$

$x = -2$

$y = \frac{2}{5}\cdot (-2)^{2}-\frac{32}{5}\cdot (-2)-\frac{72}{5}$

$y(-2) = 0$

$x = 8$

$y = \frac{2}{5}\cdot (8)^{2}-\frac{32}{5}\cdot (8)-\frac{72}{5}$

$y(8) = -40$

$r = \frac{-40-0}{10}$

$r = -4$

The average rate of change of the parabola is -4.