## Precalculus (6th Edition) Blitzer

Using $f\left( x \right)=3{{x}^{2}}+x$, we can determine that $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}=7$. This means that the point-slope equation of the tangent line to the graph of $f\left( x \right)=3{{x}^{2}}+x$ at $\left( 1,4 \right)$ is $y-4=7\left( x-1 \right)$.
Consider the function $f\left( x \right)=3{{x}^{2}}+x$ Since it is given that $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}=7$, thus the slope of the tangent line to the graph of the function is $7$. The point-slope equation of the tangent line to the graph of $f\left( x \right)=3{{x}^{2}}+x$ at $\left( 1,4 \right)$ with the slope $7$ is found as follows: Substitute the value of ${{x}_{1}}=1,{{y}_{1}}=4\text{ and }m=7$ in $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ $y-4=7\left( x-1 \right)$ Therefore, the tangent line to the graph of $f\left( x \right)=3{{x}^{2}}+x$ at $\left( 1,4 \right)$ is $y-4=7\left( x-1 \right)$.