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Find the equation of the tangent to the curve `y- x^2 + 4x - 5 = 0`   at the point...

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anniepalmer | Student, Undergraduate | (Level 1) eNoter

Posted November 6, 2012 at 3:42 AM via web

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Find the equation of the tangent to the curve

`y- x^2 + 4x - 5 = 0`   at the point (5,10)

y-X2+4x-5=o

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durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted November 6, 2012 at 4:24 AM (Answer #1)

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 Write the equation in standard form (y=)

`y= x^2 -4x + 5` To find the tangent-find the gradient at the point (5;10)   

`y'=2x - 4`  (Remember, the derivative gives you the gradient. Substitute for x as we want the gradient at that point.)

`y' = 2(5) - 4`

`y' = 6` = m (the gradient)

A tangent has an equation : y=mx+ c

We have a point (5;10) and we have m and can therefore find c:

y= mx+c becomes

`10 = (6)(5) +c`

`therefore c=-20`

`therefore` y=6x - 20

 

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eagudelo | Student, College Freshman | (Level 1) Honors

Posted November 6, 2012 at 4:17 AM (Answer #2)

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For this you have to use implicit differentiation (which is usually learned in Calculus I) :
y - x^2 +4x - 5 = 0     Differentiate
y' - 2x + 4 = 0             Solve for y'
y' = 2x - 4                   Plug in 5 for x
y' = 2(5)-4 = 10-4=6   6 is the slope of the tangent line when x = 5
Now use y = mx + b to find the actual equation. You have the slope, m, which is 6, and you have y, which is 10, and x, which is 5. So,
y = mx + b                  Substitute
10 = 6(5) + b              Solve for b
10 = 30 + b
b = 10-30 = -20          Plug b into the tangent equation. y=6x-20.

So, the equation of the tangent to the curve y - x^2 +4x - 5 = 0 is y=6x-20.
This works because in Calculus I you learn that y' is the slope of the line that is tangent to the point given. I hope this helps!

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