What is the equation of a tangent line to the curve y = x^2 which also passes thru the point (-13, 144).?

Question

??? y = -16x + 64 ???

Answer 1

y = 2x_0*x + y_0 - 2x_0^2 or y - y_0 = 2x_0(x - x_0)

Answer 2

I won't do the completed challenge (notwithstanding I in simple terms solved it myself), yet i visit get you going so which you would be able to end it your self. once you're in a calculus class, you understand the spinoff (slope) of the parabola is y' = d/dx(x^2) = 2x. So the slope of the line you're searching for is 2x, the place x is the factor of tangency. Now assume the factor of tangency is at x = a. Then y = x^2 = a^2, and the completed coordinates are (a,a^2). The slope is 2x = 2a. Now you have 2 factors -- (13,one hundred forty four) and (a,a^2) -- and the slope 2a. Given 2 factors, the slope formula is (a^2 - one hundred forty four)/(a - 13) = 2a sparkling up this for a. you will get a quadratic which you would be able to ingredient, (the factor of tangency would be at (a,a^2).) as quickly as you have a, then plug it into the factor-slope formula (y - one hundred forty four)/(x - 13) = 2a sparkling up for y and that often is the excellent answer.

Answer 3

y-144 = m(x+13)
must intersect y = x^2 and have the slope of 2x at the intersect point.
2x^2 +26x + 144 = x^2
x^2 + 26x + 144 = 0
x = (-26+/--sqrt(676-576))/2 = -13+/-5
x = (-8, -21)
Intersect points are (-8.64), and (-18, 324)

y-144 = -16(x+13)
y = -16x - 64

y-144 = -36(x + 13)
y = -36x - 324

Both lines fit the requirements.

Answer 4

A line that is tangent to y=x^2 at the point (x0,y0) has the equation

y - y_0 = m(x - x_0)

where m = y' at x = x_0. y' = 2x, so the equation becomes

y - y_0 = 2x_0(x - x_0),
or
y = 2x_0*x + y_0 - 2x_0^2.

Since y_0 = x_0^2, it simplifies to
y = 2x_0*x - x_0^2.
Now, plug in x = -13 and y = 144, and solve for x_0.

You get
144 = 2x_0(-13) - x_0^2,
or
x_0^2 + 26x_0 + 144 = 0.
This factors into
(x_0+18)(x_0+8) = 0,
so x_0 = -18 or x_0 = -8.

Plug in x_0 = -8, and you get
y = -16x - 64
If x_0 = -18, you get
y = -36x - 324.

Answer 5

Here's the original equation:

f(x) = x^2

You find the derivative:

f'(x) = 2x

Then find the derivative @ x=-13:

f'(-13) = -26

Use point-slope formula to find the equation (y-y1=m(x-x1)):

y - 144 = -26(x + 13)

Now simplify:

y = -26x - 194

James L did it WRONG, but the guy directly above me is correct.

Answer 6

f'(x)=2x (eqn of a grad)
Sub x=-13 into f'(x)
f'(-13)=-26
Grad of the tang is -26
Use Pt Grad formula y-y1=m(x-x1) Sub (-13, 144) and m=-26 into it
y-144=-26(x+13)
y=-26x-194 (eqn of the tang)