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

Answer 1

Important to note at the start: (13, 144) is NOT on the curve!

All points on this curve have coordinates (a,a^2) for some number a.
At any point (a, a^2) on the curve, the slope of the tangent line at that point is 2a.

Here's what we know about the line we are looking for:

It passes through the points (13,144) and (a, a^2)
The slope of the line is 2a.

Thus we can write the following equation using two different ways of expressing slope:
2a = (a^2 - 144)/(a - 13)

Multiplying both sides by (a - 13), we get:
2a(a - 13) = a^2 - 144
2a^2 - 26a = a^2 - 144
a^2 - 26a + 144 = 0
(a - 8)(a - 18) = 0
a = 8 or a = 18

Thus there are two tangent lines satisfying your criteria, one passing through the point (8,64) and the other passing through the point (18,324)

The line passing through (8,64) has slope 16, so the equation of that line is :
y - 64 = 16(x - 8)
or
y = 16x - 64

Similarly, the line passing through (18,324) has slope 36. Thus its equation is:
y - 144 = 36(x - 13) (I used the point (13,144) for the point-slope form because the numbers were smaller.)

or
y = 36x - 324

To summarize: The equations of the two lines are:

y = 16x - 64 and y = 36x - 324

Answer 2

if y=x^2, then the slope of the tangent line is equivalent to the derivative. y'=2x + some unknown constant. Plug the point numbers in to get 144 = 2 * 13 + unknown constant. Unknown constant equals 118.

Thus, your answer is y = 2*x+118

Answer 3

y-one hundred and forty four = m(x+13) ought to intersect y = x^2 and performance the slope of 2x on the intersect element. 2x^2 +26x + one hundred and forty four = x^2 x^2 + 26x + one hundred and forty four = 0 x = (-26+/--sqrt(676-576))/2 = -13+/-5 x = (-8, -21) Intersect factors are (-8.sixty 4), and (-18, 324) y-one hundred and forty four = -16(x+13) y = -16x - sixty 4 y-one hundred and forty four = -36(x + 13) y = -36x - 324 both lines fit the criteria.

Answer 4

use differnation on y=x^2-->dy/dx=2x.
sub x=13, dy/dx=26
use this gradient to find the equation
y-144=26(x-13)
doing the step....
u get y=26x-194

Answer 5

find y' for slope
slope = 2
(y-y1)=m(x-x1)
(y-144)=2(x-13) solve for y