Question

How do you find the equation of the tangent line to the curve `-x^2+5x` at (-1,-5)?

Answer 1

There is something wrong with this question. However, the method is shown. You will need to adjust the values I used.

Explanation:

Set `color(white)("d")y=-x^2+5x" ".................Equation(1)`

Check: Set `x=-1 => y= -1+5(-1)= -6`

`color(red)("The point (-1,-5) is NOT on the curve "y=-x^2+5)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(magenta)("Method assuming that (-1,-5) is on the curve")`

Increment `x` by the minute amount of `deltax`

As you have changed the `x` part in doing this then the `y` part will also change.

Let the resuting change in `y` be `deltay` so after the change we have:

`y+deltay=-(x+deltax)^2+5(x+deltax)" "...Equation(2)`

`y+deltay=-(x^2+2xdeltax+(deltax)^2)+5x+5deltax`

`y+deltay=-x^2-2xdeltax-(deltax)^2+5x+5deltax ....Eqn(2_a)`

Apply the subtraction `Eqn(2_a)-Eqn(1)`

`y+deltay=-x^2-2xdeltax-(deltax)^2+5x+5deltax`
`ul(ycolor(white)("dddd") = -x^2color(white)("ddddddddddd.d")+5x larr" Subtract")`
`color(white)("ddd") deltay=color(white)("dd")0x^2-2xdeltax-(deltax)^2+0x+5deltax`

Set gradient as `("change in " y)/("change in " x)->(deltay)/(deltax)`

so divide both sides by `deltax` to get gradient giving:

`(deltay)/(deltax)=-(2xdeltax)/(deltax) -(deltax)^2/(deltax)+(5deltax)/(deltax)`

`(deltay)/(deltax)=-2x-deltax+5`

`lim_(deltax->0)(deltay)/(deltax)=-2x-lim_(deltax->0)deltax+5`

`color(red)(f'(x)->dy/dx=-2x-0+5 larr" Gradient")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus for the straight line `color(blue)(y=color(red)(m)x+c)` we have `color(red)(m=-2x+5)`

It is given that the point is `P_1(x,y)=(-1,-5)`

Thus substitute `-1` for `x => m=-2(-1)+5=+7->+7/1=("change in y")/("change in x")`

So `y=mx+c color(white)("d")->color(white)("d")y=7x+c`

Substitute in the known point to determin `C`

`y=7x+c color(white)("dddd")->color(white)("dddd") -5=7(-1)+c`

`color(white)("ddddddddddddd")->color(white)("dddd")-5+7=c`

`color(white)("ddddddddddddd")->color(white)("dddddddd.d")c=2`

So the equation of the tangent at `P_1(x=-1,y=-5)" is "`

`y=7x+2`

Answer 2

Correction to the question
See the first solution for detailed method using the values as
given in the question.

Explanation:

`y=-x^2+5x`

thus gradient is `-2x+5`

Set the known point `P_1->(x,y)=(-1,y)`

Thus we have:

`y=-(-1)^2+5(-1) = -6`

Setting `P_1->(x,y)=(-1,-6)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Gradient at `x=1 -> m=-2(-1)+5 = +7` giving:

`y=7x+c`

Known point `(x,y)->-6=7(-1)+c color(white)("d")=>color(white)("d") c=+1`

Thus the equation of the tangent is: `y=7x+1`