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QuestionSimplify the expression: (x+Δx)2−9(x+Δx)+8−(x2−9x+8)Δx\frac{(x+\Delta x)^{2}-9(x+\Delta x)+8-(x^{2}-9 x+8)}{\Delta x}.

Studdy Solution

STEP 1

Assumptions1. We are given the expression (x+Δx)−9(x+Δx)+8−(x−9x+8)Δx\frac{(x+\Delta x)^{}-9(x+\Delta x)+8-\left(x^{}-9 x+8\right)}{\Delta x}. . We need to simplify this expression.

STEP 2

First, we need to expand the numerator of the given expression.(x+Δx)2−9(x+Δx)+8−(x2−9x+8)Δx=x2+2xΔx+(Δx)2−9x−9Δx+8−x2+9x−8Δx\frac{(x+\Delta x)^{2}-9(x+\Delta x)+8-\left(x^{2}-9 x+8\right)}{\Delta x} = \frac{x^2 +2x\Delta x + (\Delta x)^2 -9x -9\Delta x +8 -x^2 +9x -8}{\Delta x}

STEP 3

Now, we need to simplify the numerator by cancelling out the like terms.
x2+2xΔx+(Δx)2−9x−9Δx+8−x2+9x−8Δx=2xΔx+(Δx)2−9ΔxΔx\frac{x^2 +2x\Delta x + (\Delta x)^2 -9x -9\Delta x +8 -x^2 +9x -8}{\Delta x} = \frac{2x\Delta x + (\Delta x)^2 -9\Delta x}{\Delta x}

STEP 4

Now, we can factor out Δx\Delta x from the numerator.
2xΔx+(Δx)2−9ΔxΔx=Δx(2x+Δx−9)\frac{2x\Delta x + (\Delta x)^2 -9\Delta x}{\Delta x} = \Delta x(2x + \Delta x -9)

STEP 5

Now, we can cancel out Δx\Delta x from the numerator and the denominator.
Δx(2x+Δx−9)Δx=2x+Δx−9\frac{\Delta x(2x + \Delta x -9)}{\Delta x} =2x + \Delta x -9So, the simplified form of the given expression is 2x+Δx−92x + \Delta x -9.

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