How do you find the average rate of change of #y=x^2+6x+10# over [1,3]?

Question

Emma Aldridge · Accepted Answer

#(Delta y)/(Delta x) = 10#

Given:

#f(x) = x^2+6x+10#

The average rate of change of #f(x)# over any interval #[a, b]# is:

#(Delta f(x))/(Delta x) = (f(b)-f(a))/(b-a)#

In other words, it is the slope of the chord joining #(a, f(a))# and #(b, f(b))#

In our instance, we discover:

#f(1) = 1^2+6(1)+10 = 17#

#f(3) = 3^2+6(3)+10 = 37#

So the average rate of change of #f(x)# over #[1, 3]# is:

#(Delta f(x))/(Delta x) = (f(3)-f(1))/(3-1) = (37-17)/2 = 20/2 = 10#

Samantha Adkison · Answer

undefined To find the average rate of change of $y = x^2 + 6x + 10$ over the interval $[1, 3]$, you first evaluate the function at the endpoints of the interval, then subtract the values and divide by the change in $x$.

1. Evaluate the function at the endpoints of the interval:
   - At $x = 1$, $y = (1)^2 + 6(1) + 10 = 1 + 6 + 10 = 17$.
   - At $x = 3$, $y = (3)^2 + 6(3) + 10 = 9 + 18 + 10 = 37$.

2. Calculate the change in $y$ and the change in $x$:
   - Change in $y = 37 - 17 = 20$
   - Change in $x = 3 - 1 = 2$

3. Compute the average rate of change:
   - Average rate of change $= \frac{\text{Change in } y}{\text{Change in } x} = \frac{20}{2} = 10$.

Therefore, the average rate of change of $y = x^2 + 6x + 10$ over the interval $[1, 3]$ is $10$.