Free math worksheets

- Use only the function values at the two endpoints of the interval. - Compare the change in output with the change in input. - Keep the endpoint order the same in both differences.

1. From the table, \(f(-1)=4\) and \(f(2)=1\). 2. The average rate of change is \(\frac{f(2)-f(-1)}{2-(-1)}=\frac{1-4}{3}=-1\).

\(-1\)

- Evaluate the quadratic at each endpoint of the interval. - Form a quotient using the change in function value over the change in input. - Check that the endpoint order is consistent in the numerator and denominator.

1. Evaluate the endpoints: \(f(1)=2-3+1=0\) and \(f(4)=2\cdot4^2-3\cdot4+1=21\). 2. The average rate of change is \(\frac{f(4)-f(1)}{4-1}=\frac{21-0}{3}=7\).

\(7\)

- Evaluate the height at the beginning and end of the stated time interval. - Divide the change in height by the elapsed time. - Include units and describe what the sign of the rate means for the ball's height.

1. Evaluate the endpoint heights: \(h(0.5)=25\) and \(h(1.5)=41\). 2. The average rate of change is \(\frac{41-25}{1.5-0.5}=16\,\text{ft/s}\). 3. Over this interval, the ball's height increased by an average of \(16\) feet for each second of elapsed time.

\(16\,\text{ft/s}\). From \(0.5\) to \(1.5\) seconds, the ball's height increased at an average rate of \(16\) feet per second.

- Evaluate the function at \(0\), \(2\), and \(4\). - Compute each endpoint-to-endpoint change using the same quotient structure. - Compare both the signs and magnitudes of the two rates with the location of the vertex.

a) \(f(0)=1\) and \(f(2)=-3\), so the average rate of change is \(\frac{-3-1}{2-0}=-2\). b) \(f(2)=-3\) and \(f(4)=1\), so the average rate of change is \(\frac{1-(-3)}{4-2}=2\). c) The negative rate on \([0,2]\) shows the function decreases on average as it approaches the vertex at \(x=2\). The positive rate on \([2,4]\) shows it increases on average after the vertex. The equal magnitudes reflect the parabola's symmetry about \(x=2\).

a) \(-2\) b) \(2\) c) The function decreases toward the vertex and then increases away from it, with equal-magnitude rates on these symmetric intervals.

- Read each marked point as an ordered pair before calculating. - Average rate of change compares the vertical change with the horizontal change between the endpoints. - Once you know the slope, use either marked point to determine the secant line.

a) From the graph, \(A=(-1,1)\) and \(B=(2,-2)\). b) The average rate of change is \(\frac{-2-1}{2-(-1)}=\frac{-3}{3}=-1\). c) The secant line has slope \(-1\). Using point \(A\), \(y-1=-1(x+1)\), so \(y=-x\). The secant-line slope is exactly the average rate of change on the interval.

a) \(A=(-1,1)\) and \(B=(2,-2)\) b) \(-1\) c) \(y=-x\); its slope \(-1\) equals the average rate of change.

- Express both endpoint function values in terms of the unknown coefficient. - Use the given average rate to build an equation rather than evaluating a known function. - After solving for the coefficient, substitute it back into the original function rule.

1. Evaluate the endpoints in terms of \(a\): \(f(1)=a+1\) and \(f(3)=9a+5\). 2. Use the given average rate: \(\frac{(9a+5)-(a+1)}{3-1}=10\). 3. Simplify: \(\frac{8a+4}{2}=10\), so \(4a+2=10\) and \(a=2\). 4. Therefore, \(f(x)=2x^2+2x-1\).

\(a=2\), so \(f(x)=2x^2+2x-1\).

- Compare how far each endpoint is from the vertex's x-coordinate. - Evaluate the two endpoint expressions in vertex form before expanding anything. - Think about what equal endpoint heights imply about a secant line.

a) The endpoints are equally far from \(x=3\). Thus, \(f(3-d)=2d^2-5\) and \(f(3+d)=2d^2-5\). Their outputs are equal, so the average rate of change is \(\frac{0}{(3+d)-(3-d)}=\frac{0}{2d}=0\). b) The parabola is symmetric about the vertical line \(x=3\). Points the same horizontal distance to the left and right of the axis have equal heights, so every secant through such a symmetric pair is horizontal.

a) \(0\) b) The interval endpoints are symmetric about \(x=3\), so they have equal function values and the secant line is horizontal.

- Write the average-rate condition using the marked starting point and an unknown second endpoint. - Simplify the resulting quotient only after noting why the unknown endpoint cannot equal the starting x-value. - After finding the x-coordinate, evaluate the quadratic to locate the second endpoint on the graph.

1. Since \(f(1)=-2\), the required condition is \(\frac{f(x)-(-2)}{x-1}=3\), with \(x>1\). 2. Substitute the function: \(\frac{x^2-4x+3}{x-1}=3\). 3. Factor the numerator: \(x^2-4x+3=(x-1)(x-3)\). Because \(x>1\), \(x\neq1\), so the quotient simplifies to \(x-3\). 4. Solve \(x-3=3\), giving \(x=6\). Then \(f(6)=36-24+1=13\), so \(B=(6,13)\).

\(x=6\), and the second endpoint is \(B=(6,13)\).