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For the function
f(x)=x^(2)-10, find the slope of the secant line between
x=-6 and
x=1.
Answer:

For the function $f(x)=x^{2}-10$ , find the slope of the secant line between $x=-6$ and $x=1$ . $\newline$ Answer:

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Find equations of tangent lines using limits

Full solution

Q. For the function

f(x)=x^{2}-10

, find the slope of the secant line between

x=-6

and

x=1

.

\newline

Answer:

Calculate Function Values: To find the slope of the secant line between two points on a function, we use the formula for slope, which is the change in $y$ divided by the change in $x$ (rise over run). This is given by the formula $(f(x_2) - f(x_1)) / (x_2 - x_1)$ , where $x_1$ and $x_2$ are the $x$ -values of the two points.
Find Two Points: First, we need to find the y-values for the function at $x = -6$ and $x = 1$ . We do this by plugging these x-values into the function $f(x) = x^2 - 10$ . For $x = -6$ : $f(-6) = (-6)^2 - 10 = 36 - 10 = 26$ . For $x = 1$ : $f(1) = (1)^2 - 10 = 1 - 10 = -9$ .
Calculate Slope: Now we have the two points on the function: $(-6, 26)$ and $(1, -9)$ . We can use these to find the slope of the secant line. $\newline$ Slope = $(f(1) - f(-6)) / (1 - (-6)) = (-9 - 26) / (1 + 6) = -35 / 7 = -5$ .
Final Result: The slope of the secant line between $x = -6$ and $x = 1$ for the function $f(x) = x^2 - 10$ is $-5$ .

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