Calculus problem - slope of parabola?
Find slope of parabola y=xsq at point p (2,4) and Q [2+h, (2+h)sq]
Step 1: secant slope Delta y/delta x = [(2+h)sq - 2sq] /h
= (hsq + 4h +4 -4) /h
= (hsq + 4h) /h = h + 4
What do I do after this?
Find slope of parabola y=xsq at point p (2,4) and Q [2+h, (2+h)sq]
Step 1: secant slope Delta y/delta x = [(2+h)sq - 2sq] /h
= (hsq + 4h +4 -4) /h
= (hsq + 4h) /h = h + 4
What do I do after this?
Comments
The slope of the tangent line is the same as the slope of the secant line except, there is lim(h->0) in front of the whole equation.
So step 2 would be:
lim(h->0) (h+4)
Then using direct substitution, you get
0+4 = 4
Therefore, slope of tangent line of the parabola y=x^2 at the point(2,4) is 4.
Find slope of parabola y = x² at point P(2,4) and Q[(2 + h),(2 + h)²].
The slope, m, is given by:
m = [(y2) - (y1)]/[(x2) - (x1)]
m = [(2 + h)² - 4]/[(2 + h) - 2]
m = [4 + 4h + h² - 4]/[2 + h - 2]
m = (4h + h²)/h
m = h(4 + h)/h
m = 4 + h
-4