Question

What is the slope of the line through (2,4) and (x,y) for y = x2 + x - 2 and x=1.99? x=2.004? x=2+h.

What happens to this last slope when h is very small?

Answer 1

Part 1)
`m=4.99`

Part 2)
`m=5.004`

Part 3)
`m=h+5`

Part 4) The slope tends to 5.

Explanation:

we know that

The formula to calculate the slope between two points is equal to

`m=(y2-y1)/(x2-x1)`

we have

the points (2,4) and (x,y)

`y=x^(2) +x-2`

Part 1)

For x=1.99

substitute the value of x in the quadratic equation and solve for y

`y=(1.99)^(2) +(1.99)-2`

`y=3.9501`

we have the points

(1.99,3.9501) and (2,4)

substitute the values in the formula of slope

`m=(4-3.9501)/(2-1.99)`

`m=(0.0499)/(0.01)`

`m=4.99`

Part 2)

For x=2.004

substitute the value of x in the quadratic equation and solve for y

`y=(2.004)^(2) +(2.004)-2`

`y=4.020016`

we have the points

(2,4) and (2.004,4.020016)

substitute the values in the formula of slope

`m=(4.020016-4)/(2.004-2)`

`m=(0.020016)/(0.004)`

`m=5.004`

Part 3)

For x=2+h

substitute the value of x in the quadratic equation and solve for y

`y=(2+h)^(2) +(2+h)-2\\\\y=4+4h+h^(2) +2+h-2\\\\y=h^(2)+5h+4`

we have the points

(2,4) and (2+h,h^{2}+5h+4)

substitute the values in the formula of slope

`m=(h^(2)+5h+4-4)/(2+h-2)`

`m=(h^(2)+5h)/(h)`

Simplify

`m=h+5`

Part 4) What happens to this last slope when h is very small

`m=h+5`

If the value of h is very small, then the value of h tends to zero and the value of m tends to 5.

therefore

`m=5`