A quadratic function is graphed. Compare the estimated average rate of change for the quadratic function to the estimated average rate of change for the exponential function y=−6x

Question

on the interval [0,5]
. Use one of the symbols <
, =
, or >
to describe the comparison.

(1 point)
The estimated average rate of change of the quadratic function
the estimated average rate of change of the exponential function.

Answer 1

To compare the estimated average rate of change for the quadratic function and the exponential function \( y = -6x \) on the interval \([0, 5]\), we need to compute the average rate of change for both functions over that interval.

For the exponential function \( y = -6x \):

1. Calculate \( y(0) \): \[ y(0) = -6(0) = 0 \]
2. Calculate \( y(5) \): \[ y(5) = -6(5) = -30 \]
3. The average rate of change from \( x = 0 \) to \( x = 5 \) is given by: \[ \text{Average rate of change} = \frac{y(5) - y(0)}{5 - 0} = \frac{-30 - 0}{5 - 0} = \frac{-30}{5} = -6 \]

Now we need to estimate the average rate of change of the quadratic function over the same interval. Since you have not provided a specific quadratic function, let's denote it generically as \( f(x) = ax^2 + bx + c \). You will need to calculate its average rate of change similarly using \( f(0) \) and \( f(5) \).

Assuming \( f(0) = c \) and \( f(5) = 25a + 5b + c \), we calculate the average rate of change:

4. The average rate of change for the quadratic function is given by: \[ \text{Average rate of change} = \frac{f(5) - f(0)}{5 - 0} = \frac{(25a + 5b + c) - c}{5} = \frac{25a + 5b}{5} = 5a + b \]

Now, you can compare \( 5a + b \) with \(-6\).

• If \( 5a + b < -6 \) : the quadratic function's average rate of change is less than that of the linear function.
• If \( 5a + b = -6 \) : they are equal.
• If \( 5a + b > -6 \) : the quadratic function's average rate of change is greater than that of the linear function.

Without specific values for \( a \) and \( b \), I can't give you a definitive answer, but you can follow these steps to compare the two once you have the coefficients of your quadratic function.

Answer 2

the values are (0,0) (8,0)

Answer 3

Given that the quadratic function has its points as (0,0) and (8,0), it appears that the quadratic function is defined by the points where it intersects the x-axis at these points. The simplest form of a quadratic function that meets these criteria is \( f(x) = k(x)(x - 8) \), where \( k \) is a constant.

Step 1: Find the average rate of change of the quadratic function.

Let's assume \( k = 1 \) for simplicity, so the quadratic function is: \[ f(x) = x(8 - x) = -x^2 + 8x \]

However, we can just analyze the average rate of change between the x-values 0 and 5 since the function still passes through those points.

1. Calculate \( f(0) \): \[ f(0) = 0(8 - 0) = 0 \]
2. Calculate \( f(5) \): \[ f(5) = 5(8 - 5) = 5 \times 3 = 15 \]
3. The average rate of change of the quadratic function from \( x = 0 \) to \( x = 5 \) is: \[ \text{Average rate of change} = \frac{f(5) - f(0)}{5 - 0} = \frac{15 - 0}{5 - 0} = \frac{15}{5} = 3 \]

Step 2: Recall the average rate of change of the exponential function.

From the previous calculation:

• The average rate of change for the exponential function \( y = -6x \) over the interval \([0, 5]\) was calculated to be \(-6\).

Step 3: Compare the two average rates of change.

• Average rate of change of the quadratic function: \( 3 \)
• Average rate of change of the exponential function: \( -6 \)

Final Comparison:

Now, we can compare the two rates of change: \[ 3 > -6 \]

Thus, we conclude that:

The estimated average rate of change of the quadratic function is greater than the estimated average rate of change of the exponential function.

The comparison can be expressed as: \[ 3 > -6 \]

In the context of your question: The estimated average rate of change of the quadratic function > the estimated average rate of change of the exponential function.