Question

a. for the function $f(x)=-6x^{2}+6x + 7$ find $m_{sec}=$, the slope of the secant line between the values $x_1$ and $x_2$. (round to 6 decimal places.)
a. $x_1 = 3$ and $x_2 = 4$, $m_{sec}=square$
b. $x_1 = 3$ and $x_2 = 3.1$, $m_{sec}=square$
c. $x_1 = 3$ and $x_2 = 3.01$, $m_{sec}=square$
b. use the answers from parts a,b,c to estimate $m_{tan}$, the slope of the tangent line to $f$ at point $x = 3$. (the answer should be an integer). $m_{tan}=square$

Explanation:

Step1: Recall secant - slope formula

The slope of the secant line between two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the function $y = f(x)$ is given by $m_{sec}=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. First, find $f(x)=-6x^{2}+6x + 7$.

Step2: Calculate $f(x_1)$ and $f(x_2)$ for part a

For $x_1 = 3$, $f(x_1)=-6\times3^{2}+6\times3 + 7=-6\times9 + 18+7=-54 + 18+7=-29$.
For $x_2 = 4$, $f(x_2)=-6\times4^{2}+6\times4 + 7=-6\times16+24 + 7=-96+24 + 7=-65$.
Then $m_{sec}=\frac{f(x_2)-f(x_1)}{x_2 - x_1}=\frac{-65+29}{4 - 3}=\frac{-36}{1}=-36.000000$.

Step3: Calculate $f(x_1)$ and $f(x_2)$ for part b

For $x_1 = 3$, $f(x_1)=-29$ (calculated above).
For $x_2 = 3.1$, $f(x_2)=-6\times(3.1)^{2}+6\times3.1 + 7=-6\times9.61+18.6 + 7=-57.66+18.6 + 7=-32.06$.
Then $m_{sec}=\frac{f(x_2)-f(x_1)}{x_2 - x_1}=\frac{-32.06 + 29}{3.1 - 3}=\frac{-3.06}{0.1}=-30.600000$.

Step4: Calculate $f(x_1)$ and $f(x_2)$ for part c

For $x_1 = 3$, $f(x_1)=-29$ (calculated above).
For $x_2 = 3.01$, $f(x_2)=-6\times(3.01)^{2}+6\times3.01 + 7=-6\times9.0601+18.06+7=-54.3606+18.06 + 7=-29.3006$.
Then $m_{sec}=\frac{f(x_2)-f(x_1)}{x_2 - x_1}=\frac{-29.3006 + 29}{3.01 - 3}=\frac{-0.3006}{0.01}=-30.060000$.

Step5: Estimate the slope of the tangent line

As $x_2$ gets closer to $x_1 = 3$, the slope of the secant line approaches the slope of the tangent line. From the values $m_{sec}=-36.000000,-30.600000,-30.060000$, we estimate $m_{tan}=-30$.

Answer:

a. -36.000000
b. -30.600000
c. -30.060000
B. -30