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For the function
f(x)=x^(2)+2x-5, find the slope of the secant line between
x=-3 and
x=5.
Answer:

For the function $f(x)=x^{2}+2 x-5$ , find the slope of the secant line between $x=-3$ and $x=5$ . $\newline$ Answer:

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Complex conjugate theorem

Full solution

Q. For the function

f(x)=x^{2}+2 x-5

, find the slope of the secant line between

x=-3

and

x=5

.

\newline

Answer:

Calculate $f(-3)$ : 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 $x$ , or $(f(x_2) - f(x_1)) / (x_2 - x_1)$ . We need to calculate the function values at $x = -3$ and $x = 5$ .
Calculate $f(5)$ : First, calculate $f(-3)$ . Substitute $x$ with $-3$ into the function $f(x) = x^2 + 2x - 5$ . $\newline$ $f(-3) = (-3)^2 + 2(-3) - 5 = 9 - 6 - 5 = -2$ .
Use slope formula: Next, calculate $f(5)$ . Substitute $x$ with $5$ into the function $f(x) = x^2 + 2x - 5$ . $\newline$ $f(5) = (5)^2 + 2(5) - 5 = 25 + 10 - 5 = 30$ .
Use slope formula: Next, calculate $f(5)$ . Substitute $x$ with $5$ into the function $f(x) = x^2 + 2x - 5$ . $f(5) = (5)^2 + 2(5) - 5 = 25 + 10 - 5 = 30$ .Now, use the slope formula with $f(-3)$ and $f(5)$ to find the slope of the secant line. $\text{Slope} = \frac{f(5) - f(-3)}{5 - (-3)} = \frac{30 - (-2)}{5 - (-3)} = \frac{30 + 2}{5 + 3} = \frac{32}{8} = 4$ .

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