Using the function f(x) = X^2 + 10 and the point (2,14) answer the following:

Question

(a) Find the slope of the graph of f at the given point.

(b) Find an equation of the tangent line to the graph at the point.

It would help a lot if you could show your work. Thanks!

Answer 1

Using the function f(x) = X^2 + 10 and the point (2,14) answer the following:

(a) Find the slope of the graph of f at the given point.

(b) Find an equation of the tangent line to the graph at the point.

As y= x² +10

dy/dx = slope = d/dx(x² +10)

= 2x

Slope at (2,14)

=2*2 = 4

Let the equation of tangent be

y= mx +c

y= 4x+c

as it passes through (2,14)

14= 4*2 +c gives c= 6

Hence equation of tangent

y=4x+6

Answer 2

(a) There is a long way to do this and a short way, both of which are calculus (not precalculus). If you have learned the shortcut methods, the derivative of x^2 is 2x and the derivative of 10 is 0. So the slope formula for F(x) = x^2 + 10 is 2x + 0 or 2x. Then substitute the point value for x so 2x = 2(2) = 4.

(b) The slope of the tangent line is 4. You can find the equation of the line two different ways (Slope intercept or point slope). Most students prefer using y = mx + b. Then we already know that the slope m = 4.

So we have y = 4x + b. To find b, substitute the point (2, 14) for x and y.

This means 14 = 4(2) + b.Then 14 = 8 + b. So b = 6. The equation of the tangent line is y = 4x + 6.