Question

Find the slope of the tangent line to the graph of fat the given point.$f\left(x\right)=1+2x-3x^2,at(1,0)$.

Short answer

The slope of the tangent line is $-4$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$\begin{array}{l}f\left(x\right)=1+2x-3x^2\\ po\mathrm{int}=(1,0)\end{array}$

Step 2. Formula used.

The tangent line to the curve$y=f\left(x\right)$at the point$P(a,f(a\left)\right)$is the line through p with slope,

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

provided that this limit exists.

Step 3. Finding the slope of tangent line at given point.

Let$f\left(x\right)=1+2x-3x^2$,then the slope of the tangent line at$(1,0)$is,

$\begin{array}{c}m=\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{(1+2x-3x^2)-(1+2(1)-3{\left(1\right)}^2)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{1+2x-3x^2-(3-3)}{x-1}\\ =\underset{x\to 1}{\lim }\frac{-(3x+1)\cancel{(x-1)}}{\cancel{(x-1)}}\\ =\underset{x\to 1}{\lim }-(3x+1)\\ =-(3\cdot 1+1)\\ =-4\end{array}$

Therefore, the slope of the tangent line is -4.