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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)=x2+2x−5 f(x)=x^{2}+2 x-5 , find the slope of the secant line between x=−3 x=-3 and x=5 x=5 .\newlineAnswer:

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Full solution

Q. For the function f(x)=x2+2x−5 f(x)=x^{2}+2 x-5 , find the slope of the secant line between x=−3 x=-3 and x=5 x=5 .\newlineAnswer:
  1. Calculate f(−3)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 yy divided by the change in xx, or (f(x2)−f(x1))/(x2−x1)(f(x_2) - f(x_1)) / (x_2 - x_1). We need to calculate the function values at x=−3x = -3 and x=5x = 5.
  2. Calculate f(5)f(5): First, calculate f(−3)f(-3). Substitute xx with −3-3 into the function f(x)=x2+2x−5f(x) = x^2 + 2x - 5.\newlinef(−3)=(−3)2+2(−3)−5=9−6−5=−2f(-3) = (-3)^2 + 2(-3) - 5 = 9 - 6 - 5 = -2.
  3. Use slope formula: Next, calculate f(5)f(5). Substitute xx with 55 into the function f(x)=x2+2x−5f(x) = x^2 + 2x - 5.\newlinef(5)=(5)2+2(5)−5=25+10−5=30f(5) = (5)^2 + 2(5) - 5 = 25 + 10 - 5 = 30.
  4. Use slope formula: Next, calculate f(5)f(5). Substitute xx with 55 into the function f(x)=x2+2x−5f(x) = x^2 + 2x - 5.f(5)=(5)2+2(5)−5=25+10−5=30f(5) = (5)^2 + 2(5) - 5 = 25 + 10 - 5 = 30.Now, use the slope formula with f(−3)f(-3) and f(5)f(5) to find the slope of the secant line.Slope=f(5)−f(−3)5−(−3)=30−(−2)5−(−3)=30+25+3=328=4\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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