Answer
a) The slope of the tangent to the graph of $f\left( x \right)=5{{x}^{2}}$ at $\left( -2,20 \right)$ is $-20$.
b) The slope intercept equation of the tangent line is $y=-20x-20$.
Work Step by Step
(a)
Consider the function $f\left( x \right)=5{{x}^{2}}$ and the point $\left( -2,20 \right)$
Here, $\left( a,f\left( a \right) \right)=\left( -2,20 \right)$
Now, compute the slope of the tangent line for the function $f\left( x \right)=5{{x}^{2}}$ as follows:
Put $a=-2$ ,
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( -2+h \right)-f\left( -2 \right)}{h}$
To compute $f\left( -2+h \right)$, substitute $x=-2+h$ in the function $f\left( x \right)=5{{x}^{2}}$.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 5{{\left( -2+h \right)}^{2}} \right]-5{{\left( -2 \right)}^{2}}}{h}$
Now, simplify ${{\left( -2+h \right)}^{2}}$ by using the property ${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$.
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{5\left( 4-4h+{{h}^{2}} \right)-20}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{20-20h+5{{h}^{2}}-20}{h}
\end{align}$
Combine the like terms in the numerator.
${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{5{{h}^{2}}-20h}{h}$
Now, factor the numerator and divide the numerator as well as the denominator by h,
$\begin{align}
& {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{5h\left( h-4 \right)}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,5\left( h-4 \right)
\end{align}$
Apply the limits and simplify.
$\begin{align}
& {{m}_{\tan }}=5\left( 0-4 \right) \\
& =-20
\end{align}$
Thus, the slope of the tangent to the graph of $f\left( x \right)=5{{x}^{2}}$ at $\left( -2,20 \right)$ is $-20$.
(b)
Consider the function $f\left( x \right)=5{{x}^{2}}$ and the point $\left( -2,20 \right)$
From part (a), the slope of the tangent to the graph of $f\left( x \right)=5{{x}^{2}}$ at $\left( -2,20 \right)$ is $-20$.
Now compute the slope intercept equation of the tangent line as follows:
Here, the tangent line passes through the point $\left( -2,20 \right)$ which implies ${{x}_{1}}=-2$ and ${{y}_{1}}=20$
Substitute $m=-20,{{x}_{1}}=-2,{{y}_{1}}=20$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.
$\begin{align}
& y-20=-20\left[ x-\left( -2 \right) \right] \\
& y-20=-20\left( x+2 \right) \\
& y=-20x-40+20 \\
& y=-20x-20
\end{align}$
Thus, the slope intercept equation of the tangent line is $y=-20x-20$.
