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Find the slope of the curve y=x2-5x-4 at the point P(3,-10) by finding the limit of the secant slopes through point P. Can you please show me how to solve?

Question

Find the slope of the curve y=x2-5x-4 at the point P(3,-10) by finding the limit of the secant slopes through point P. Can you please show me how to solve?

check_circleAnswer
Step 1

The given equation is y(x) = x2 – 5x – 4 and the given point is (3,–10).

Let the other point of intersection of the secant and the curve be

(3 + h, y (3 + h)).

Substitute x(3+h) in y(x)=x-5x-4
y(3+h)(3+h)-5(3+h)-4
- h2 +h-10
Thus, the other point is (3+h,h h-10)
help_outline

Image Transcriptionclose

Substitute x(3+h) in y(x)=x-5x-4 y(3+h)(3+h)-5(3+h)-4 - h2 +h-10 Thus, the other point is (3+h,h h-10)

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Step 2
It is known that, the straight line passes through two points (x,.y) and
y2-y1
(x2 y then the slope of the line is m
=
(h+h-10)-(10)
(3+h)-3
m=
h2h1010
3 h 3
h
-h+1
It is known that, the slope of curve at (3,-10) is equal to limit of secant
slopes
That is lim(h1) 1
h-0
help_outline

Image Transcriptionclose

It is known that, the straight line passes through two points (x,.y) and y2-y1 (x2 y then the slope of the line is m = (h+h-10)-(10) (3+h)-3 m= h2h1010 3 h 3 h -h+1 It is known that, the slope of curve at (3,-10) is equal to limit of secant slopes That is lim(h1) 1 h-0

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