Question

The point P(3,18) lies on the curve y = x2 + x + 6. If Q is the point (x,x2 + x + 6), find the slope of the secant linePQ for the following values of x. Ifx = 3.1,
the slope of PQ is: and if x = 3.01,
the slope of PQ is: and if x = 2.9,
the slope of PQ is: and if x = 2.99,
the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(3,18).

Answer 1

Answer:

x =3.1

$m = \frac{3.1^2 +3.1+6 -18}{3.1-3} = 7.1$

x =3.01

$m = \frac{3.01^2 +3.01+6 -18}{3.01-3} = 7.01$

x =2.9

$m = \frac{2.9^2 +2.9+6 -18}{2.9-3} = 6.9$

x =2.99

$m = \frac{2.99^2 +2.99+6 -18}{2.99-3} = 6.99$

Based on the results we can conclude this condition:

$y'(3) = 7$

And the slope would be given by:

$y'(x) = 2x+1$

Step-by-step explanation:

For this case we have defined the point P=(3,18) and Q= (x, x^2 +x+6) and we want to find the slope of the secant for the line PQ for a list of values.

The slope for the secant would be given by this formula:

$m = \frac{Q_y -P_y}{Q_x -P_x}$

And if we replace the values for the points we got:

$m = \frac{x^2 +x+6 -18}{x-3}$

So now we just need to replace the different values for x

x =3.1

$m = \frac{3.1^2 +3.1+6 -18}{3.1-3} = 7.1$

x =3.01

$m = \frac{3.01^2 +3.01+6 -18}{3.01-3} = 7.01$

x =2.9

$m = \frac{2.9^2 +2.9+6 -18}{2.9-3} = 6.9$

x =2.99

$m = \frac{2.99^2 +2.99+6 -18}{2.99-3} = 6.99$

Based on the results we can conclude this condition:

$y'(3) = 7$

And the slope would be given by:

$y'(x) = 2x+1$