Question

Write the equation of the tangent to the graph of the function
`y=x^3-3x-2`at the point x0=1

Answer 1

`y=-x-3`

Explanation:

Differentiation is an algebraic process that finds the gradient (slope) of a curve. At a point, the gradient of a curve is the same as the gradient of the tangent line to the curve at that point. Therefore, the gradient of the equation of the tangent line to function f(x) at x = c is f'(c).

Differentiate the given function.

`\begin{aligned}y&=x^2-3x-2\\\\\implies \frac{\text{d}y}{\text{d}x}&=2x-3\end{aligned}`

To find the slope, m, of the tangent line at x = 1, substitute x = 1 into the differentiated function:

`\implies m=2(1)-3=-1`

Substitute x = 1 into the original function to find the y-coordinate of the point on the curve at x = 1.

`\implies y= (1)^2-3(1)-2=-4`

Substitute the found slope m = -1 and point (1, -4) into the point-slope formula to create an equation of the tangent line at x = 1:

`\implies y-y_1=m(x-x_1)`

`\implies y-(-4)=-1(x-1)`

`\implies y+4=-x+1`

`\implies y=-x-3`

Therefore, the equation of the tangent to the graph of the given function at x = 1 is:

`\boxed{y=-x-3}`

`\hrulefill`

Differenation Rules

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=nx^(n-1)$\\\end{minipage}}`

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\frac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}`

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating a constant}\\\\If $y=a$, then $\frac{\text{d}y}{\text{d}x}=0$\\\end{minipage}}`