Y = x^2 − 7x/x^2 + 4x

at (2, − 5/6)
(a) At the indicated point, find the slope of the tangent line.

(b) At the indicated point, find the instantaneous rate of change of the function.

a) The slope of the line tangent to the curve that passes through the point (2,-10) is equal to the derivative of p at x=2.

Using differentiation rules (power rule and sum rule), the derivative of p(x) for any x is p'(x)=3x^2-9. In particular, the value we are looking for is p'(2)=3(2^2)-9=12-9=3.

If you would like to compute the equation of the tangent line, we can use the point-slope equation to get y=3(x-2)-10=3x-16

b) The instantaneus rate of change is also equal to the derivative of P at the point x=2, that is, P'(2). This is equal to p'(2)=3.

Step-by-step explanation:

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

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