Question

If the tangent to the curve $y=x^3+ax+b$ at $(1,-6)$ is parallel to the line $x-y+5=0$, find $a$ and $b$.

Answer 1

Step 1:Differentiate the given equation of the curve with respect to $x$

It is given that the tangent to the curve is $y=x^3+ax+b$ at point $(1,-6)$ and parallel line $x-y+5=0$.

We have to find the slope of tangent.

Let,

$y=x^3+ax+b$ ------ $\left(1\right)$

$x-y+5=0$ -------- $\left(2\right)$

We have to differentiate the equation $y=x^3+ax+b$ with respect to $x$.

So,

$\Rightarrow$ $\frac{dy}{dx}=\frac{d}{dx}\left(x^3+ax+b\right)$

Solve by using the Sum/Difference rule, $\left(f\pm g\right)\text{'}=f\text{'}\pm g\text{'}$.

$\Rightarrow$ $\frac{dy}{dx}=\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(ax\right)+\frac{d}{dx}\left(b\right)$

$\Rightarrow$ $\frac{dy}{dx}=3x^{3-1}+a\left(\frac{dx}{da}\right)+0$

$\Rightarrow$ $\frac{dy}{dx}=3x^2+a$

Step 2: Find the slope of the tangent to the curve $y=x^3+ax+b$ at $(1,-6)$.

The first order derivative of the equation of a curve at a given point gives the slope of the tangent at that point

We will find the slope of the tangent to the curve $y=x^3+ax+b$ at point $(1,-6)$ is,

$\Rightarrow$ ${\frac{dy}{dx}}_{\left(1,-6\right)}=3{\left(1\right)}^2+a$

$\Rightarrow$ ${\frac{dy}{dx}}_{\left(1,-6\right)}=3+a$ ---------- $\left(3\right)$

Step 3: Find the slope of the line $x-y+5=0$

From the given, parallel line is $x-y+5=0$.

The equation $x-y+5=0$ is the form of equation of a straight line $y=mx+c$. Here $m$ is the slope of line.

Therefore, the slope of the line is

$y=1\times x+5$

Therefore, the slope is $1$.

Step 4: Compare the slopes to find the value of $a$

The slopes of parallel lines are equal

From equation $\left(3\right)$,

$\Rightarrow$ $3+a=1$

$\Rightarrow$ $a=-2$

Step 5: Using equation of given curve find the value of $b$

So, point $(1,-6)$ lies on the curve, that is the given points are satisfies the equation $\left(1\right)$.

We will put $x=1$ and $y=$ $-6$ and $a=-2$ in equation $\left(1\right)$. We get,

$\Rightarrow$ $-6=1^3+\left(-2\right)\left(1\right)+b$

$\Rightarrow$ $-6=1-2+b$

$\Rightarrow$ $b=-5$

Therefore, the value of $a$ is $-2$ and $b$ is $-5$.