## Precalculus (6th Edition) Blitzer

a) The slope of the tangent to the graph of $f\left( x \right)={{x}^{2}}+4$ at $\left( -1,5 \right)$ is $-2$. b) The slope intercept equation of the tangent line is $y=-2x+3$.
(a) Consider the function $f\left( x \right)={{x}^{2}}+4$ and the point $\left( -1,5 \right)$ Here, $\left( a,f\left( a \right) \right)=\left( -1,5 \right)$ Now, compute the slope of the tangent line for the function $f\left( x \right)={{x}^{2}}+4$ as follows: Put $a=-1$ , ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( -1+h \right)-f\left( -1 \right)}{h}$ To compute $f\left( -1+h \right)$, substitute $x=-1+h$ in the function $f\left( x \right)={{x}^{2}}+4$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( -1+h \right)}^{2}}+4 \right]-\left[ {{\left( -1 \right)}^{2}}+4 \right]}{h}$ Now, simplify ${{\left( -1+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 1-2h+{{h}^{2}}+4 \right]-5}{h}$ Combine the like terms in the numerator. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}-2h}{h}$ Now, factor the numerator and divide the numerator as well as the denominator by h, \begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h-2 \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( h-2 \right) \end{align} Apply the limits and simplify. \begin{align} & {{m}_{\tan }}=0-2 \\ & =-2 \end{align} Thus, the slope of the tangent to the graph of $f\left( x \right)={{x}^{2}}+4$ at $\left( -1,5 \right)$ is $-2$. (b) Consider the function $f\left( x \right)={{x}^{2}}+4$ and the point $\left( -1,5 \right)$ From part (a), the slope of the tangent to the graph of $f\left( x \right)={{x}^{2}}+4$ at $\left( -1,5 \right)$ is $-2$. Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point $\left( -1,5 \right)$ which implies ${{x}_{1}}=-1$ and ${{y}_{1}}=5$ Substitute $m=-2,{{x}_{1}}=-1,{{y}_{1}}=5$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. \begin{align} & y-5=-2\left[ x-\left( -1 \right) \right] \\ & y-5=-2\left( x+1 \right) \\ & y=-2x-2+5 \\ & y=-2x+3 \end{align} Thus, the slope intercept equation of the tangent line is $y=-2x+3$.