The point P(191) lies on the curve y=82 x^2+9
(a) If Q is the point (x,82 x^2+9), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x.
If x=05 then the slope equals : ?
Thank you
(a) If Q is the point (x,82 x^2+9), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x.
If x=05 then the slope equals : ?
Thank you
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P(1,91) and Q(x, 82x²+9)
Slope of PQ:
(82x²+9 - 91) / (x-1)
= (82x² - 82) / (x-1)
= (82(x-1)(x+1)) / (x-1)
= 82(x+1)
If that's x = 5, then the slope is 82(5+1) = 492
If it's another value of x, just plug it in to 82(x+1)
Slope of PQ:
(82x²+9 - 91) / (x-1)
= (82x² - 82) / (x-1)
= (82(x-1)(x+1)) / (x-1)
= 82(x+1)
If that's x = 5, then the slope is 82(5+1) = 492
If it's another value of x, just plug it in to 82(x+1)
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You're welcome!
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