Finding the slope of a line

Question

I am trying to help my daughter do her calculus homework. It's been 30 years since I did this stuff, so it's greek to me. Any help would be appreciated.

suppose f(x)=14-x^2. find the slope of the tangent line to the grapy y=F(x) when x=-2

Please show how this is achieved/worked so I can explain/show it to her so she will know how to work a similar problem. Thanks.

Answer 1

f(x)= 14-x^2

f'(x) =-2x

at the point where x =-2

f'(-2) = -2(-2) = 4 = slope of the tangent line at the point where x=-2

slope = the derivative

derivative of 14-x^2 = the derivative of 14 minus the derivative of x^2

derivative of 14 or any constant = 0

that leaves the derivative of x^2

power rule is make the exponent the coefficient and reduce the exponent by 1

f'(x) =-2x^(2-1) = -2x

replace x by -2 to find the derivative when x=2

f'(-2) = -2(-2) = 4

you might also graph the function with a graphing calculator, online or handheld

it's a downward opening parabola with vertex (0,14)= maximum point.

It's symmetric about the vertical line x=0 which is the y axis

for x=-2, you're on the left side of the parabola with a somewhat steep slope, slope =4 looks about right

or you can do the quotient difference as h approaches zero

f(x) =14-x^2

f(x+h) = 14-(x+h)^2 = 14-x^2 -2xh -h^2

f(x+h) - f(x) = -2xh-h^2

divide by h

= -2x -h

let h= 0

= -2x = the derivative = slope of the tangent line

replace x with -2

= -2(-2) = 4

Answer 2

The derivative is the slope of the tangent line.

If f(x) = 14 - x² then, using the power rule, the derivative is given as f '(x) = -2x

At x = -2, f '(-2) = -2(-2) = 4

So the slope of the tangent line at x = -2 is 4

Note: The power rule is a shortcut to finding the derivative of terms that have a variable base and a number exponent. It is that the derivative of xⁿ = n•x^n-1