1. (1 point) The point P(5;34) lies on the curve y=x2+x+4. Let Q be the point (x

Question

1. (1 point) The point P(5;34) lies on the curve y=x2+x+4.
Let Q be the point (x;x2+x+4).
a.) Compute the slope of the secant line PQ for the following
values of x:
When x = 5:1 , the slope of PQ is:
When x = 5:01 , the slope of PQ is:
When x = 4:9 , the slope of PQ is:
When x = 4:99 , the slope of PQ is:
b.) Based on the above results, guess the slope of the tangent
line to the curve at P(5;34).

Spring2018 1 Functions and Models: Problem 11 (s poinc) The Sgures bekow s uncthion b(z) has y antescept 1 and goes through the point (o,a + 1) show the graphs of the (a) Find a formuia for fa) hep (formua (c) Find a tomuia for h(z) help formulas

Explanation / Answer

a)

When x = 5:1

slope of PQ =[f(5.1) -f(5)]/(5.1-5) =(35.11 -34)/(0.1) =11.1

When x = 5:01 ,

slope of PQ =[f(5.01) -f(5)]/(5.01-5) =(34.1101 -34)/(0.01) =11.01

When x = 4.9 ,

slope of PQ =[f(4.9) -f(5)]/(4.9-5) =(32.91-34)/(4.9-5) =10.9

When x = 4.99 ,

slope of PQ =[f(4.99) -f(5)]/(4.99-5) =(33.8901-34)/(4.99-5)  =10.99

b)

From the above result, when x is approaching 5 (from left or right), the slope of secant line approaches to value 11.

hence, he slope of the tangent line to the curve at P(5;34) is 11.

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