Finding the what? Tangent line? What the heck is that....

Hey, just started college and I don't understand a word that comes out of my proffessor's mouth.
We have to turn in homework online and I tried finding the tangent line for this equation, but the the answer came out wrong, and I don't want to risk fixing the answers and using up the last chance I have to submit the correct answer.

Find the equation of the line tangent to y=(x^4)+9(x^2)-x at the point (1,9)

My first answer was y=23x-14, but it came out to be wrong. Anyone mind telling me what the heck I did wrong and explaining in detail what the correct way of solving this problem is.

First, determine the slope of the line by finding the derivative of the function at the point of tangency.

y = x^4 + 9x^2 - x
y' = 4x^2 + 18x - 1
y'(1) = 4 + 18 -1 = 21

The slope of the line is 21, and the tangent point is a point on the line. Use the point-slope formula for a line to determine the equation of the line:

(y-9)/(x-1) = 21
y-9 = 21(x-1)

Standard form of the equation: 21x - y = 12
Slope-intercept form: y = 21x - 12
Intercept form: x/(12/21) + y/(-12) = 1

Pretty sure this is done using derivatives-- these can be thought of as taking the n power and multiplying by the base, up to the n-1. So;

y=(x^4)+9(x^2)-x
= 4x^3 + 9*2x + 1x^0
slope at pt 1;
m = 4(1)^3 + 18(1) + 0
= 4+18 = 22

y - 9 = 22(x-1)
= y = 22x +8

y = x^4 + 9x^2 - x

dy/dx = 4x^3 + 18x - 1

At (1, 9), m = 21.

y - 9 = 21(x - 1)
y = 21x - 21 + 9
y = 21x - 12