Determine line perpendicular to tangent of the curve y=-3x^2 + 1 at the point (2,11). Can be done without der?

Question

Can it be done without the derivative or whatever that is?

Answer 1

No, we need derivative.

Slope of the tangent line at given point:

y' = -6x , so at x = 2, this is -12.

therefore, slope of the perpendicular line is 1/12.

So, we have the slope, 1/12, and a point (2,11), the line is

(y-11)/ (x-2) = 1/12 or

y-11 = x/12 - 2/12 or
y-11 = x/12 - 1/6 or
y = x/12 + 10 5/6 , in slope-intercept form.

Answer 2

the only way to find the tangent of the curve at a point is to find its derivative, NO OTHER WAY!
y= f(x) = - 3x^2 + 1
slope at x =2 (where y = -11, not 11)
= f'(x) at x =2
f'(x) = -6x +1
f'(2) = -11

the line perpendicular will have slope 1/11
at (2, -11), the equation is
(y + 11)/(x - 2) = 1/11
11( y + 11 ) = x - 2
11y + 121 = x - 2
11y - x + 123 = 0

Answer 3

Step a million: what's the slope of "y=a million" --> slope m=0. Step 2: locate the by-made of f(x), and locate all factors with slope m=0: f(x) = ln^2(x^2) f'(x) = 2x (2 ln x^2) (a million/x^2) = 4/x (ln x^2) = 0 ln x^2 = 0 x^2 = a million x = -a million or a million Step 3: now plug in to locate the factors: x = a million, f(x) = ln^2(a million^2) = 0 x = -a million, f(x) = ln^2((-a million)^2) = 0 So the factors are (a million,0) and (-a million,0).

Answer 4

1) the point is (2,-11)
yes.
Take a line through (2-11)
y+11= m(x-2) and intercept with y=-3x^2+1
-3x^2+12 =mx-2m so 3x^2 +mx-2m-1 2= 0 This equation must
have a double root
so m^2+12(2m+12)=0 m2+24m+144=0 so (m+12)^2=0
and m=-12
The tangent is y+11= -12(x-2)

Answer 5

your the science guy, you figure it out.