slope of secant lines
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for the function ( f(x)=-2 x^{2}-3 x - 10 ), find the slope of the secant line between ( x=-4 ) and ( x = 1 ).
answer attempt 1 out of 2

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Accepted Answer

# Explanation:
## Step1: Find $ f(-4) $
Substitute $ x = -4 $ into $ f(x)=-2x^{2}-3x - 10 $.
$$
\begin{align*}
f(-4)&=-2\times(-4)^{2}-3\times(-4)-10\
&=-2\times16 + 12-10\
&=-32+12 - 10\
&=-30
\end{align*}
$$
## Step2: Find $ f(1) $
Substitute $ x = 1 $ into $ f(x)=-2x^{2}-3x - 10 $.
$$
\begin{align*}
f(1)&=-2\times1^{2}-3\times1-10\
&=-2-3-10\
&=-15
\end{align*}
$$
## Step3: Use the slope formula for the secant line $ m=\frac{f(x_2)-f(x_1)}{x_2 - x_1} $
Here $ x_1=-4,x_2 = 1,f(x_1)=-30,f(x_2)=-15 $
$$
\begin{align*}
m&=\frac{-15-(-30)}{1-(-4)}\
&=\frac{-15 + 30}{1 + 4}\
&=\frac{15}{5}\
&=3
\end{align*}
$$
# Answer:
$3$

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QUESTION IMAGE

Question

Explanation:

Step1: Find \( f(-4) \)

Step2: Find \( f(1) \)

Step3: Use the slope formula for the secant line \( m=\frac{f(x_2)-f(x_1)}{x_2 - x_1} \)

Answer: