Letf(x)=πx4−8x3+2x−11.

Question

Determine an equation of the tangent line to the graph of f at x = −2.

Determine an equation of the line that passes through the graph of f at x = −2, but is perpendicular to the tangent line to the graph of f at x = −2. (This line is called a normal line to the graph of f.)

Answer 1

First, we should look for the slope of the tangent line for f(-2). To find the slope of an equation at any given point, we can take the derivative of the function to obtain an equation with which we can find this slope. Remember that, to take the derivative of exponents, you bring the exponent down as a coefficient of the exponentiated variable/expression, subtract 1 from the exponent, and then multiple this by the derivative of the exponentiated expression. We don't have any expressions in parentheses, so that last part just means we technically multiply the term by 1, but since that doesn't do anything, I will omit that. Also, the derivative of a constant (-11) is 0. The derivative would be:

f '(x) = π(4)x^3 - 8(3)x^2 + 2

= 4πx^3 - 24x^2 + 2

To find the slope at x = -2, we plug in -2:

f '(-2) = 4π(-2)^3 - 24(-2)^2 + 2

= 4π(-8) -24(4) + 2

= -32π - 96 + 2

= -32π - 94 = -2(16π + 47)

Now that we have the slope of the tangent line, we can obtain the slope of the normal line. The slope of a line perpendicular to a given line is the opposite reciprocal of the given line's slope. Therefore the slope of the normal line would be

-1/(-2(16π + 47))

I can pull out a -1 from that denominator, so I can simplify this to be:

-1/(-1(2)(16π + 47)) = 1/(2(16π + 47))

The slope of a line is y = mx + b. We know the slope of our normal line is 1/(32π+94), let's plug it in:

y = x/(2(16π + 47)) + b

To figure out what our y-intercept is, we need to input a point on this line and solve for it. The only point we know is on this line is the point where it intersects with the original function. We know that the x value is -2, but we do not know the y value. Let's solve for it:

f(-2) = π(-2)^4 - 8(-2)^3 + 2(-2) - 11

= π(16) - 8(-8) - 4 - 11

= 16π + 64 - 15

= 16π + 49

Now we know that the point on the original function where this normal line touches is (-2,16π + 49), we can plug this into the normal line's equation and solve for b:

16π + 49 = -2/(2(16π + 47)) + b

16π + 49 = -1/(16π+47) + b

16π + 49 + 1/(16π+47) = b

((16π + 49)(16π + 47) + 1)/(16π+47) = b

((256π^2 + 752π + 784π + 2303) + 1)/(16π+47) = b

(256π^2 + 1536π + 2303 + 1)/(16π+47) = b

(256π^2 + 1536π + 2304)/(16π+47) = b

256(π^2 + 6π + 9)/(16π + 47) = b

256(π+3)^2/((16π + 47) = b

Now let's plug this back into our equation for the normal line:

y = x/(2(16π + 47)) + 256(π+3)^2/(16π + 47)

You could write the answer as this, though I like to combine terms, so:

y = x/(2(16π + 47)) + 2*256(π+3)^2/(2(16π + 47)

y = (x+512(π+3)^2)/(2(16π + 47))

Hope that helps! I checked my work and graphed these functions, but I am human, so if anyone catches a mistake, please comment!

Answer 2

I will not crunch the numbers for you, but I will tell you how to do it.

First, substitute x=-2 into the expression for f(x) to get a value...call it y.

Differentiate f(x) and evaluate the derivative at x=-2; this is the slope of the tangent line...call it m.

Now use -2,y and m along with the point-slope form to get the equation of the tangent.

The slope of the normal line is -1/m.

Use the point-slope form again to get the equation of the normal.