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

Aug 12, 2018

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 $x = - 1 \implies y = - 1 + 5 \left(- 1\right) = - 6$

$\textcolor{red}{\text{The point (-1,-5) is NOT on the curve } y = - {x}^{2} + 5}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{m a \ge n t a}{\text{Method assuming that (-1,-5) is on the curve}}$

Increment $x$ by the minute amount of $\delta x$

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

Let the resuting change in $y$ be $\delta y$ so after the change we have:

$y + \delta y = - {\left(x + \delta x\right)}^{2} + 5 \left(x + \delta x\right) \text{ } \ldots E q u a t i o n \left(2\right)$

$y + \delta y = - \left({x}^{2} + 2 x \delta x + {\left(\delta x\right)}^{2}\right) + 5 x + 5 \delta x$

$y + \delta y = - {x}^{2} - 2 x \delta x - {\left(\delta x\right)}^{2} + 5 x + 5 \delta x \ldots . E q n \left({2}_{a}\right)$

Apply the subtraction $E q n \left({2}_{a}\right) - E q n \left(1\right)$

$y + \delta y = - {x}^{2} - 2 x \delta x - {\left(\delta x\right)}^{2} + 5 x + 5 \delta x$
ul(ycolor(white)("dddd") = -x^2color(white)("ddddddddddd.d")+5x larr" Subtract")
$\textcolor{w h i t e}{\text{ddd") deltay=color(white)("dd}} 0 {x}^{2} - 2 x \delta x - {\left(\delta x\right)}^{2} + 0 x + 5 \delta x$

Set gradient as $\left(\text{change in " y)/("change in } x\right) \to \frac{\delta y}{\delta x}$

so divide both sides by $\delta x$ to get gradient giving:

$\frac{\delta y}{\delta x} = - \frac{2 x \delta x}{\delta x} - {\left(\delta x\right)}^{2} / \left(\delta x\right) + \frac{5 \delta x}{\delta x}$

$\frac{\delta y}{\delta x} = - 2 x - \delta x + 5$

${\lim}_{\delta x \to 0} \frac{\delta y}{\delta x} = - 2 x - {\lim}_{\delta x \to 0} \delta x + 5$

$\textcolor{red}{f ' \left(x\right) \to \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x - 0 + 5 \leftarrow \text{ Gradient}}$
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Thus for the straight line $\textcolor{b l u e}{y = \textcolor{red}{m} x + c}$ we have $\textcolor{red}{m = - 2 x + 5}$

It is given that the point is ${P}_{1} \left(x , y\right) = \left(- 1 , - 5\right)$

Thus substitute $- 1$ for $x \implies m = - 2 \left(- 1\right) + 5 = + 7 \to + \frac{7}{1} = \left(\text{change in y")/("change in x}\right)$

So $y = m x + c \textcolor{w h i t e}{\text{d")->color(white)("d}} y = 7 x + c$

Substitute in the known point to determin $C$

$y = 7 x + c \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} - 5 = 7 \left(- 1\right) + c$

$\textcolor{w h i t e}{\text{ddddddddddddd")->color(white)("dddd}} - 5 + 7 = c$

$\textcolor{w h i t e}{\text{ddddddddddddd")->color(white)("dddddddd.d}} c = 2$

So the equation of the tangent at ${P}_{1} \left(x = - 1 , y = - 5\right) \text{ is }$

$y = 7 x + 2$ Aug 12, 2018

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

#### Explanation:

$y = - {x}^{2} + 5 x$

thus gradient is $- 2 x + 5$

Set the known point ${P}_{1} \to \left(x , y\right) = \left(- 1 , y\right)$

Thus we have:

$y = - {\left(- 1\right)}^{2} + 5 \left(- 1\right) = - 6$

Setting ${P}_{1} \to \left(x , y\right) = \left(- 1 , - 6\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Gradient at $x = 1 \to m = - 2 \left(- 1\right) + 5 = + 7$ giving:

$y = 7 x + c$

Known point $\left(x , y\right) \to - 6 = 7 \left(- 1\right) + c \textcolor{w h i t e}{\text{d")=>color(white)("d}} c = + 1$

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