Problem 868

Find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y  = 7 – 4x2 ,

P (-2,-9).




a)    Start with a secant line through P(-2,-9) and Q(-2+h, 7-4(-2+h)2) nearby.  Write an expression for the slope the secant PQ, and determine what happens to the slope as Q approaches P, that is , as h approaches 0.

Secant slope = \frac{\Delta y}{\Delta x}= \frac{(7-4(-2+h)^{2})-(7-4(-2)^{2})}{h}

Simplify the numerator.

Secant slope  = \frac{\Delta y}{\Delta x}=  \frac{16h - 4h^{2}}{h}

Now simplify the entire fraction to get 16 – 4h. Notice that this expression approaches 16 as h approaches 0.

Thus , the slope of the curve at P(-2,-9) is 16.


b)    Use this slope and the given point (-2,-9) to write the equation for the tangent line.

y – (-9) = 16(x – (-2)) This is the point-slope equation.

y = 16x + 32 – 9

y = 16 x + 23

The equation of the line tangent to y = 7 – 4x2 at P (-2,-9) is y = 16x + 23



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