# 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

*print*Print*list*Cite

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**

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!